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3rd Prep – 1st Term Step by Step Sheets Answers.

Unit One: Force and Motion.Lesson One: Motion in One Direction.

Sheet (1): Part One: Motion.1. Complete the following statements:

1. The concept of the motion is linked to the change of an object's position as time passes according to the location of another object.

2. The motion in a straight line in one direction is considered the simplest type of motion.3. The path of a moving object may be straight, Curved or combination of each.

2. Write the scientific term for each of the following: 1. The change in the position of a body by the time relative to the position of another body. Motion.2. The simplest type of motion. Motion in straight line in one direction.

3. Give reasons for: 1. The motion of trains can be considered as a motion in one direction.Because trains move in a straight line forward or backward but it doesn't move upward or downward.

Lesson One: Motion in One Direction.Sheet (2): Part Two: Speed.

1. Complete the following statements: 1. If car (A) takes a smaller time than car (B) to cover the same distance therefore, car A is faster than car B.2. The result of multiplying a speed of a moving object by time = distance.3. Speed measurement units are m/sec. or km/hr.4. A car covers 80 meters in 4 seconds so, it moves at speed equals 20 m/sec.

2. Write the scientific term for each of the following: 1. A physical quantity which is used to describe and measure the motion of objects. Speed.2. The distance that a moving object covers within a unit time. Speed.The distance moved through a unit time.

3. The result of dividing the distance over time. Speed.4. The measuring unit of the speed. m/sec. or km/hr.

3. Give reasons for: 1. Cars and planes are provided by speedometer.To help us in identifying the speed of cars and planes directly.

2. The object's speed increases by decreasing the time needed to cover a certain distance.

Because speed equals distanctime so, speed is inversely proportional to the time.

3. The speed of a moving object increases as the covered distance increases at constant time.

Because speed equals distanctime so, speed is directly proportional to the distance.

4. What is meant by each of the following: 1. An object moving in a straight line covers a distance of 20 meters in one second.The speed of the object is 20m/sec.

2. A moving car covers a distance of 100km. in two hours.The speed of the car is 50km/hr.

5. What are the two factors that can be used to describe the body motion?1. The distance that is covered by the moving body. 2. The time taken by the moving body to cover this distance.

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Lesson One: Motion in One Direction. Sheet (3): Part Three: Types of Speed.

1. Complete the following statements: 1. A train which travels a distance of 360 km. with regular speed 120 km/hr. needs 3 hours to cover this distance.2. The object which covers a distance 72 km. in an hour, its speed equals 20 m/sec.3. When a car moves with uniform speed equals 30m/sec., this means that the car covers 30 meters after one second and 120 meter after 4 seconds.

4. When an object covers equal distances at unequal periods of time so, it moves with irregular speed.2. Write the scientific term for each of the following:

1. An instrument used in cars and planes to identify speed directly. Speedometer.2. A moving object covers equal distances at equal periods of time. Regular (Uniform) speed.The speed in which the object moves to cover equal distances at equal periods of time.

3. The speed of a moving body that covers equal distances at unequal time span. Irregular speed.3. Put ( ) or ( ), then correct what is wrong:

1. Speedometer in cars is used to identify the direction of cars. () Speed.2. If a moving object covers equal distances at equal time span so, it moves with regular speed. ()3. When a moving object covers equal distances at equal periods of time so, it moves with uniform acceleration.() Speed.

4. A car moves at regular speed equals 20m/sec., this means that the car covers 60 meter in the third second. () 20.

5. A car moves with regular speed covers 500 meter in 25 sec. its speed is 200 m/sec. () 20.6. When the average speed of a moving body equals its regular speed, the motion is described as irregular motion.() Regular

7. A moving car covers equal distances at unequal periods of time so, it moves with irregular speed. ()4. Give reasons for:

1. We say that an object moves with regular speed.Because the object covers equal distances at equal periods of time.

2. Car (A) which covers 50 meters in 5 seconds is faster than car (B) which covers 64 meter in 8 seconds.

Because the speed of car (A) = 505 = 10m/sec., while that of car (B) =

648 = 8m/sec.

3. It is hard to measure regular speed for a car practically.Most of moving cars cannot move inside crowded towns all the time by uniform speed.Due to the conditions of the road.

4. We say that an object moves with irregular speed.Because the object covers equal distances at unequal periods of time or it covers unequal distances at equal

5. An object moves with irregular speed.When the object covers equal distances at unequal periods of time (or unequal distances at equal periods of time).

Lesson One: Motion in One Direction.Sheet (4): Part Four: Average Speed.

1. Complete the following statements: 1. A car moves with irregular speed, its average speed = total distance ÷ total time.2. The result of dividing the total distance that a moving object covers by the total time taken to cover this distance is average speed.

3. Average speed represents the regular speed by which the moving object moves to cover the same distance at the same time taken to cover this distance.

4. When the average speed of a moving body equals the regular speed, the motion is described as regular motion.5. A train travels from Cairo to Banha a distance of 45 km. in 1.5 hr. so, its average speed equals 30 km/hr.




2. Write the scientific term for each of the following: 1. The total distance that a moving object covers divided by the total time taken to cover this distance.Average speed.

2. It represents regular speed by which the moving object moves to cover the same distance at the same time taken to cover this distance. Average speed.

Compare between:

3. Regular speed and average speed.

P.O.C. Regular speed Average speed

Definition: It is the change of object's position by equal distances at equal periods of time.

It is the total distance covered by the moving object divided by the total time taken to cover this distance.

It equals: v=∆d∆ t

v=d totalt total

Its measuring unit: m/sec. or km/hr. m/sec. or km/hr.

Lesson One: Motion in One Direction.Sheet (5): Part Five: Relative Speed.

1. Complete the following statements: 1. A red car moves on a road at speed 80 km/hr. and a yellow car moves in the same direction at speed 70 km/hr. so, the passenger in yellow car observes the red car moves with a speed 10 km/hr.

2. The measuring of relative speed depends on the position of the observer.3. The speed of the moving object relative to the observer is relative speed.2. Write the scientific term for each of the following:

1. The speed of moving object relative to the observer. Relative speed.3. Put ( ) or ( ), then correct what is wrong:

1. Average speed is the speed of a moving object relative to the observer. () Relative.2. The value of a car's speed relative to an observer standing on the ground is the same to an observer in a moving car. () Differs from that to.

3. Measuring the relative speed for a moving car depends on the presence of speedometer which determines the speed value. () The position of an observer who.

4. The relative speed of two moving bodies in the same direction equals the sum of their speeds. ()Difference.4. Give reasons for: A moving car seems to be at rest relative to the rider of another moving car beside it with the same velocity and direction.Because the relative speed equals the difference between their speeds equals zero.




Lesson One: Motion in One Direction.Sheet (6): Part Six: Problems.

1. Complete the missing parts in the table:

Speed (m/sec.) Distance (m) Time (Sec.)20 100 55 50 108 96 12

2. Problems: 1. A runner covers 450 meters in 45 seconds. Find his speed.

d=450m. t=45 sec . v=?v=dt=450

45=10m /sec .

2. A train travels from Cairo to Alexandria a distance of 250km. in 2 hours, find its speed.

d=250km. t=2hr . v=?v=dt=250

2=125 km /hr .

3. A bus covers a distance of 120km. with a speed 90km/hr., then it covers 105km. at 70km/hr. Calculate the time needed to cover the whole distance.

d1=120 km .v1=90kmhr

.d2=105 km .v2=70kmhr

.

t 1=d1

v1=120

90=4

3hr .t 2=

d2

v2=105

70=1.5hr .t total=t 1+t 2=

43+1.5=2 5

6hr .=2hr .∧50min .

4. A car moving at a uniform speed = 80km/hr. Find its speed in m/sec.

v=80kmhr

. v=? msec

.v=80×100060×60

=22.22m /sec .

5. A driver pressed on the brake to stop a car moves with speed 20m/sec. Calculate the time taken by the car to stop after moving 200m.

v=20msec

.t=? d=200m.t=dv=200

20=10 sec .

6. One of the foxes attacked a deer at rest. If the speed of the fox is 120km/hr., calculate the distance (by km. and meters) covered by the fox to catch the deer if the time taken is 10sec.

v=120 kmhr

.d=? t=10 sec .v=120×100060×60

=33.33m/ sec .d=v ×t=33.33×10=333.3m=0.3333 km.

7. Calculate the average speed of a car moves in a circular path its length 300 m for 10 rounds in 3 minutes.

Total distance=200×10=3000m .v=dTotaltTotal

= 30003×60

=16.67 m/sec.




8. A body moves in a straight line and the distance covered in different times are recorded in the following table:

Distance (m) 10 20 30 40Time (sec.) 5 10 15 20

a. Calculate the body speed.

v=dt=10

5=2m /sec .

b. What is the type of the speed?

v=dt=10

5=20

10=30

15=40

20=2m /sec .It's regular speed.

9. A boy on a bike covers 300 meter in a minute and 420 meter in the following minute. Calculate its average speed during the:

a. First minute.

v=d totalt total

= 30060

=5m /sec .

b. Second minute.

v=d totalt total

= 42060

=7m /sec .

c. Two minutes together.

v=d totalt total

= 300+42060+60

=6m /sec .

Lesson One: Motion in One Direction.Sheet (7): Part Six: Problems.

1. Problems: 1. Study the following figure, then answer the following:

a. Does the car move at regular speed? Why?No, because the car covers unequal distances at equal periods of time.

b. Calculate the car speed during:1. The first second.

v=dt=20

1=20m / sec.

2. The third second.

v=dt=30

1=30m / sec.

c. What is the average speed of the car during the three seconds together?

v=d totalt total

= 20+25+303

=753

=25m / sec .

2. A racer covered a distance of 100 meters of a straight track in 10 seconds then, he returned back walking, he took 80 seconds to come back to the starting point of running. Calculate the racer's average speed:




a. While running.

v=d totalt total

=10010

=10m /sec .

b. While returning.

v=d totalt total

=10080

=1.25m /sec .

c. During the whole trip.

v=d totalt total

=100+10010+80

=2.2m /sec .

3. A racer ends the race in 2 hours with average speed 25km/hr., if you know that the racer covers the first 20km. in an hour; calculate the speed of the racer after the first hours to complete the race in the given time.d total=v× t total=25×2=50 km.The remainingdistance=50−20=30km.The remaining time=2−1=1hr .

The racer speed after the first hour=dt=30

1=30 km /hr .

4. A body calculate its average speed on a bike covers 300 meters in a minute and 420 meters in the next minute. Calculate its average speed:

a. During the first minute.

d=300m, t=1×60=60 sec .v=d totalt total

=30060

=5m/sec.

b. During the second minute.

d=420m ,t=1×60=60 sec .v=d totalt total

= 42060

=7m/sec.

c. Within the two minutes.

Total distance=300+420=720mTotal time=2×60=120 sec.v=d totalt total

=720120

=6m/sec.

5. Two trains move parallel to each other but in the opposite direction, the speed of the first train is 60km/hr. and the speed of the second is 90km/hr. Calculate the relative speed of the first that observed by passengers in the second train.Relative speed=90+60=150 km/hr .

6. Car (A) moves with speed 60km/hr. and car (B) moves in the same direction with speed 90km/hr. Find the relative speed of car (B) relative to an observer:

a. Stand on the ground.Relative speed of car (B )¿anobserver on the ground=90km /hr .

b. In car (A).Relative s peed of car (B ) ¿anobserver∈car (A )=90−60=30km /hr .




Unit One: Force and Motion.Lesson Two: Graphic Representation of Moving in a Straight Line.

Sheet (1): Part One: Importance of Using Graphs.1. Complete the following statements:

1. Physicists use another mathematical relation like Graphs and Tables to predict the relation between certain physical quantities.

2. Give reasons for: 1. Physicists use mathematical relations like graphs and tables in many physical phenomena.In order to:

a. Predict the relation between certain physical quantities. b. Understand practical results. c. Describe physical phenomena in an easier way.

Lesson Two: Graphic Representation of Moving in a Straight Line.Sheet (2): Part Two: Graphic Representation of Regular Speed.

1. Complete the following statements: 1. The (distance - time) graph of an object moves at uniform speed is represented by a Straight line passing through the Origin point.

2. At regular motion, the distance is Directly proportional with Time3. The (speed - time) graph for motion at uniform speed is represented by a Straight line parallel to the Time axis.

4. The opposite graph shows the relation between the speed and time for a car moves at Regular speed.

2. Write the scientific term for each of the following: 1. The graph a regular motion at uniform speed which is represented by a straight line passing through the origin point. The (distance - time) graph.

2. The graph for a regular motion at uniform speed which is represented by a straight line parallel to the (x) axis.The (speed - time) graph.

3. Put ( ) or ( ), then correct what is wrong: 1. The (distance – time) graph for an object moves at regular speed is represented by a straight line parallel to the (Y) axis. () passing through the origin point.

2. The distance covered by a body moving at regular speed is inversely proportional to time needed to cover this distance. () directly

3. The (speed – time) graph for a moving object at regular speed is represented by a straight line parallel to the speed axis. () time




Lesson Two: Graphic Representation of Moving in a Straight Line.Sheet (3): Part Three: Acceleration.

1. Complete the following statements: 1. When the distance is measured by meter and time by seconds therefore the measuring unit of speed is m/sec. while the measuring unit of acceleration is m/sec 2 .

2. Write the scientific term for each of the following: 1. The amount of change in the body’s speed in a second. Acceleration.2. The change of speed per unit time. Acceleration.3. The measuring unit of acceleration. m/sec 2 or km/hr 2

3. Give reasons for: 1. The body which moves with acceleration can’t moves at a regular speed.Because its speed changes by time.

4. Define each of the following: 1. Acceleration.It is the change of an object's speed in one second.

5. Compare between: 1. Speed and acceleration (definition and measuring unit).

P.O.C. Speed AccelerationDefinition: It is the distance moved through unit time It is the change of an object's speed in one secondMeasuring unit: m/sec. or km/hr. m/sec2 or km/hr2.

Lesson Two: Graphic Representation of Moving in a Straight Line.Sheet (4): Part Four: Types of Acceleration.

1. Complete the following statements: 1. When the car’s speed increases, the movement is described as Accelerating motion, while when the car’s speed decreases the movement is described as Decelerating motion.

2. If the body moves from rest so, its initial speed equals Zero.3. Acceleration is the change of object’s Speed in one Second.

4. Acceleration (a) = Speed (∆v )Time (∆ t)

.

5. Acceleration increases if the object’s speed Increases by time, while it decreases if the object’s Speed decreases.6. When an object moves with decelerating motion, this means that its Initial speed is greater than its Final speed.7. Uniform acceleration is the change of object’s Speed by equal values through Equal periods of time.

2. Write the scientific term for each of the following: 1. The change of the object’s speed by equal values through equal periods of time. Uniform acceleration.2. A kind of motion in which the initial speed of a moving object is greater than its final speed. Decelerating motion.

3. A kind of motion in which the object’s speed increases by time. Accelerating motion.3. Put ( ) or ( ), then correct what is wrong:

1. Decreasing motion means that the object’s speed decreases by time. ()2. Acceleration is the change of an object’s distance in one second. () speed

3. Acceleration (a) = Initial speed−Final speed

∆ t . () Final speed−Initial speed

∆ t .

4. When a body moves at acceleration (-3) m/sec2, this means that its initial speed is greater than its final speed.()

4. Give reasons for: 1. When the driver of a moving car uses the break, we describe the car’s movement as decelerating motion.Because the car's speed decreases.

2. The acceleration is positive when its value increases.Because when acceleration increases its final speed is greater than its initial speed.




3. The object which moves at uniform speed, its acceleration equals zero.Because there is no change in object's speed.

4. The object which its motion is accelerating, its speed is irregular.Because the body's speed changes by unequal values through equal periods of time..

5. Compare between: 1. Uniform speed and uniform acceleration.

P.O.C. Uniform speed Uniform accelerationDefinition: The speed by which the moving object

covers equal distances at equal periods of time.

It is the change of the object's speed by equal values through equal periods of time.

Measuring unit: m/sec. or km/hr. m/sec2 or km/hr2.2. Accelerating motion and decelerating motion.

P.O.C. Accelerating motion Decelerating motionDefinition: The object's speed increases by time. The object's speed decreases by time.Initial and final speed:

Initial speed < final speed. Initial speed > final speed.

Lesson Two: Graphic Representation of Moving in a Straight Line.Sheet (5): Part Five: Graphic Representation of Regular Acceleration.

1. Complete the following statements: 1. The graph for an object moving with regular acceleration is represented by Speed on the vertical axis and Time on the horizontal axis.

2. Draw the (speed – time) graph which represents: 1. A body moves with regular speed 60km/hr.2. A body moves with regular acceleration 15m/sec2.3. A body moves with zero acceleration.

(1) (2) (3)

3. Study the following graphs, then answer:

a. Describe the motion of the body in each of the following graphs:




(A) (B) (C)1. (A) The body is at rest. 2. (B) The body moves with decelerating motion. 3. (C) The body moves with accelerating motion then its speed is constant. (Acceleration = Zero).

4. Which of the following graphs represents the movement of an object at:1. Increasing acceleration.2. Decreasing acceleration.3. Zero acceleration.

(A) (B) (C)1- (A) 2- (C) 3- (B)

Lesson Two: Graphic Representation of Moving in a Straight Line.Sheet (6): Part Six: Problems.

1. Problems: 1. A car starts moving from rest on a straight road. After 4 seconds the speed becomes 12m/sec. calculate the acceleration of the car.v1=0m /sec . v2=12m / sec. ∆ t=4 sec.

a=v2−v1

∆ t=12−0

4=3m /sec2 .

2. A special car can move from rest and its speed reaches 90m/sec. in 10 seconds. What is the acceleration with which the car moved?v1=0m /sec . v2=90m /sec . ∆ t=10 sec.

a=v2−v1

∆ t=90−0

10=9m / sec2 .

3. On a straight line there is a moving bus whose speed changes from 6m/sec. to 12m/sec. during a period of 3 seconds. What is the amount of acceleration?v1=6m /sec . v2=12m / sec. ∆ t=3 sec .

a=v2−v1

∆ t=12−6

3=2m/ sec2.

4. A car moves with a speed of 100 m/sec. If the driver used the brakes to decrease the speed, so it decreases by 4 m/sec2. Calculate its speed after 10 seconds from using the brakes.




v1=100m/sec. v2=? a=−4 m/sec2 ∆ t=10 sec.

a=v2−v1

∆ t−4=

v2−10020

v2−100=−40 v2=60m/sec.

5. A train moves with uniform speed 20 m/sec. When the driver uses the brakes, the train stops after 4 sec. Calculate the acceleration by which the train moves and the type of acceleration.

v1=20 m/sec. v2=0 t=4 sec .a=v2−v1

∆ t=−20

4=−5 m/sec2.

6. An object moves with initial speed equals 10m/sec., if you know that the acceleration by which the object moves equals 3m/sec2. Calculate its final speed after 5 seconds.v1=10m /sec . a=3m /sec2 . ∆ t=5 sec.

3=v2−10

5v2=(3×5 )+10=25m /sec .

7. A train moves by a velocity 20m/sec., with uniform deceleration -2m/sec2 if the brake is applied. Find the time to stop the train.v1=20m /sec . v2=0m /sec . a=−2m /sec2 .

a=v2−v1

∆ t−2=0−20

∆ t∆ t=−20

−2=10 sec .

Lesson Two: Graphic Representation of Moving in a Straight Line.Sheet (7): Part Six: Problems.

1. Problems: 1. Within 2.5 seconds the speed of a car increases from 60m/sec. to 65m/sec., while a bike moves from rest and its speed reaches 5m/sec. in one second. Which of them moved at a greater acceleration?Car v1=60m /sec . v2=65m /sec . ∆ t=2.5 sec.

acar=v2−v1

∆ t=65−60

2.5=2m /sec2 .Bike v1=0m /sec . v2=5m /sec . ∆ t=1 sec .

abike=v2−v1

∆ t=5−0

1=5m /sec 2 .The bike moves with a greater acceleration.

2. A driver used the brake to stop the car moved by 20m/sec., if the time needed for the car to stop is 10 seconds.a. Calculate the acceleration of the moving car.v1=20m /sec . v2=0m /sec . ∆ t=10 sec.

a=v2−v1

∆ t=0−20

10=−2m /sec2 .

b. What is the type of this acceleration?Decelerating motion3.From the opposite graph which represents the motion of a train, answer the following questions:

a. What is the maximum speed of the train?30m/sec.

b. Mention the kind of motion in part (BC).Uniform (constant) speed.

c. When does the driver start to use the break?at 2.5 second (at point C)

d. Which part represents:1. Accelerating motion.(AB)

2. Decelerating motion.(CD)




4.Chart shows the contrast how to change the speed of the body for a time.Calculate:

a. The distance travelled by the body during the first four seconds.v=10 m/sec. t=4 sec .d=v ×t=40m .

b. The maximum speed reached by the body during its movement.vmax=30 m/sec.

c. Acceleration in which the body moves in the last our seconds. What is its kind?v1=30 m/sec. v2=0

∆ t=12−8=4 sec.a=v2−v1

∆t=−30

4=−7.5 m/sec2.

It is a decelerating motion.3. A driver used the brakes to stop the car moved by 20 m/sec, calculate the time taken by the car to stop, given that the acceleration of the body equals -1 m/s2.v1=20 m/sec. v2=0a=−1m/s2.

a=v2−v1

∆ t∆ t=

v2−v1

a=0−20

−1=20sec .




Unit One: Force and Motion.Lesson Three: Physical Quantities; Scalars and Vectors.

Sheet (1): Part One: Physical Quantities.1. Complete the following statements:

1. Speed is a scalar physical quantity which is measured in m/sec. or km/hr.2. Time is measured in second or hour, while mass is measured in kilogram.3. Acceleration is a vector physical quantity which is measured in m/sec 2 . 4. Force, velocity and acceleration are from the vector quantities.5. The displacement is considered as vector quantity, while the density is considered as scalar quantity.6. Force is considered vector physical quantity and mass is considered scalar physical quantity.

2. Give reasons for: 1. It is necessary to deal with physical quantities and mathematical relationships.To understand the physical phenomena which represent the greatest part of physics.

Lesson Three: Physical Quantities; Scalars and Vectors.Sheet (2): Part Two: Types of Physical Quantities.

1. Complete the following statements: 1. All physical quantities are classified into two types which are scalar and vector.2. Scalar quantity is the quantity that identifies it and is enough to identify magnitude only.3. Vector quantity is the quantity that identifies it accurately and is necessary to identify its magnitude as well as direction.

4. The vector quantity can be identified by magnitude and direction.2. Write the scientific term for each of the following:

1. The physical quantity that has magnitude only. Scalar quantity.2. The physical quantities that include time, length and mass. Scalar physical quantity.3. The physical quantity that has magnitude and direction. Vector quantity.

3. Give reasons for: 1. Velocity and force are vector quantities.Because they have magnitude and direction.




Lesson Three: Physical Quantities; Scalars and Vectors.Sheet (3): Part Three: Distance and Displacement.

1. Complete the following statements: 1. Displacement is the length of shortest straight line between two positions.2. Distance is the actual length of object’s path from the starting point to the end one.3. Distance is a scalar quantity which is measured in meters, while displacement is a vector quantity which is measured in meters also.

4. The displacement equals the distance, when the object moves in a direct straight line and in one direction.2. Write the scientific term for each of the following:

1. The distance covered in a certain direction between the starting point and the ending point. Displacement.2. The length of the shortest straight line between primary position and final position. Displacement.3. The actual length of the path that a moving object covers from the starting point to the ending point of the motion.Distance.

4. The measuring unit of displacement. Meter.3. Give reasons for:

1. Distance is a scalar quantity, while displacement is a vector quantity.Because distance is determined by magnitude only, while displacement is determined by magnitude and direction.

2. When an object moves, where its start point is the end point, its velocity is zero.Because its displacement is zero.

Lesson Three: Physical Quantities; Scalars and Vectors.Sheet (4): Part Four: Speed and Velocity.

1. Complete the following statements: 1. Speed is the rate of change of distance, while velocity is the rate of change of displacement.2. Velocity is a vector physical quantity and is measured in m/sec or km/hr.3. If the measuring unit of the displacement is meter and that of time is second so the measuring unit of the velocity is m/sec.

4. Velocity is the value of displacement at a unit time and it is a vector quantity.2. Write the scientific term for each of the following:

1. The vector quantity which is measured in meter/sec. Velocity.2. The rate of change of displacement. Velocity.3. The measuring unit of velocity. m/sec. or km/hr.4. The displacement covered in one second. Velocity.




Lesson Three: Physical Quantities; Scalars and Vectors.Sheet (5): Part Five: Problems.

1. Complete the following statements:

1. Average velocity = DisplacementTotal time .

2. When an object moves from point (A) to point (B) in a direct straight line to cover a distance 60m. in 5 sec., so the object speed equals 12m/sec, while its average velocity equals 12m/sec.

3. The plane which flies against the wind direction takes more time and consumes more fuel than that which flies Give reasons for:

1. The amount of consumed fuel by a plane flies between two cities is differ according to the wind direction.Because the plane which flies against the wind direction consumes more fuel and more time than that which flies in the same wind direction.

2. Pilots take in consideration the velocity of the wind.Because the time and the amount of consumed fuel depend on the direction of the wind.

2. Problems: 1. If you move a distance of 5 meters northward and your colleague moves a distance of 5 meter southward, compare between:

a. The distance that you covered and the distance that your colleague covered.Both distances are equal in magnitude.

b. The displacement that you covered and the displacement that your colleague covered.The two displacements are equal in magnitude and opposite in direction.

3. From the opposite figure:What is the displacement of the car after time equals:

a. 2 second.Displacement = velocity × time = 10 × 2 = 20m.

b. 5 second.Displacement = velocity × time = 10 × 5 = 50m.

4. A car covered 600 in eastward direction within 60 sec. Find its average velocity.

Displacement=600m time=60 sec v=DisplacementTTotal

=60060

=10 m/sec.

5. If a body moved from point (a) to the point (c) passing by the point (b) as shown in the figure. Calculate:

a. The distance covered by the body.Distance=20+15=35m

b. The displacement done by the body.Displacement=25m∈t hedirection A⃗C




Lesson Three: Physical Quantities; Scalars and Vectors.Sheet (6): Part Five: Problems.

1. Problems: 1. The displacement that covered by a moving body through different times are recorded in the following table:

Displacement (m) 10 20 30 40 50 60Time (sec.) 5 10 15 20 25 30

a. Represent the relation graphically.

b. Calculate the velocity from the graph.

v=60−2030−10

=4020

=2 m/sec.

2. From the opposite figure a person starts moving from a point and returns back to the start point through b, c & d.Calculate:

a. Average speed.

v=dTotalT Total

=20+40+20+4025+45+25+45

=120140

=0.86 m/sec.

b. Velocity (Give reason).Velocity = Zero, because the displacement equals zero.

3. In the opposite figure:A body starts its motion from point (A) to the south to point (B), it covers of 40 meters in 10 seconds, then it directs east to point (C) at a distance of 30 meters from point (B) in 10 seconds.Calculate:

a. The amount of body displacement equals …………………Displacement = A⃗C=50m.

b. The total time spent by the body equals …………………Total time = 10 + 10 = 20 sec.

c. The length of the total distance covered by the body equals …………………Total distance = 40 + 30 = 70 m.

d. Average velocity equals …………………

v=Displacemen tT Total

=5020

=2.5 m/sec. in the direction A⃗C




4. A car starts motion from point (A) and covers 40 meter northward to point (B) within 20sec., then 80 meter eastward to point (C) within 20 second and then 40 meter southward to the point (D) within 10 seconds. Find:

a. The total distance covered by this car.Total distance = 40 + 80 + 40 = 160 m.

b. The total time that car took to cover this distance.Total time = 20 + 20 + 10 = 50 sec.

c. The displacement from the start point to the end point.Displacement = 80 m. in the direction A⃗D

d. The average velocity of this car.

v=dTotalT Total

= 8020+20+10

=1.6 m/sec. in the direction A⃗D

5. In the opposite figure:The perimeter (circumference) of this circle 44 meters and the diameter is 14 meters. When an object moves from point (A) to point (B) to point (C) in 10 seconds.

a. The distance = 22 meters.b. The displacement = 14 meters in A⃗C direction.c. The velocity = ………………… m/sec.

v=DisplacementTime

=1410

=1.4 m/sec. in the direction A⃗C .




Unit Two: Light Energy.Lesson One: Mirrors.Sheet (1): Part One: Light Reflection.

1. Complete the following statements: 1. The phenomena of the light bouncing off in the same medium when it meets the reflecting surface is called light reflection

2. Write the scientific term for each of the following: 1. The change in the direction of light ray in the same medium, when it falls on a reflecting surface.The rebounding of the light to the same side when it strikes a reflecting surface. Light reflection.

2. Angle of incidence = Angle of reflection. First law of light reflection.3. The incident light ray, the reflected light ray and the normal from the surface of reflection at the point of incidence, all lie in one plane perpendicular to the reflecting surface. Second law of light reflection.

3. Give reasons for: 1. When you look at a mirror, you see your face image.Due to light reflection.

2. The perpendicular incident light ray on a plane mirror reflects on itself.The incident light ray falling perpendicular on a reflecting surface reflects on itself.Because the angle of incidence and the angle of reflection equal zero.

3. of a light ray that is incident on a plane mirror with an angle 45o.4. Problems:

1. If the angle between the incident ray and the reflected ray is 140o, find the angle of incidence and the angle of reflection.What is the relation between them?

The angle of incidence = the angle of reflection = 140

2 = 70o

2. If the angle between the incident light ray and the reflected light ray on a plane mirror equals 120o. Calculate the angle of incidence.

The angle of incidence = the angle of reflection = 120

2 = 60o

5. Mention the two laws of light reflection?First law: angle of incidence = angle of reflection.Second law: the incident light ray, the reflected light ray and the normal at the surface of reflection at the point of incidence all lie in one plane perpendicular to the reflecting surface.

6. Study the following figure and answer the following questions: 1. The value of angle of reflection is 55 o . 2. The value of the straight angle is 180 o . 3. The concept of angle of incidence is equal to the angle of reflection.

Lesson One: Mirrors.Sheet (2): Part Two: Concepts Concerning Reflection of Light.

1. Complete the following statements: 1. In the opposite figure:

a. The angle 2 represents angle of incidence.b. The angle 3 represents angle of reflection.

2. When a light ray falls on a reflecting surface, the angle between the incident ray and the reflecting surface is 35o, therefore the angle of reflection equals 55 o and the angle between the incident ray and the reflected ray equals 110 o .

3. When a light ray falls perpendicular on a reflecting surface, its angle of reflection equals zero.2. Write the scientific term for each of the following:

1. The light ray that falls on the reflecting surface. The incident ray.




2. The light ray that bounces from the reflecting surface. The reflected ray.3. The angle between the incident light ray and the line perpendicular on the reflecting surface. Angle of incidence.4. The angle between the reflected light ray and the line perpendicular on the reflecting surface from the point of incidence. Angle of reflection.

Lesson One: Mirrors.Sheet (3): Part Three: Plane Mirror.

1. Complete the following statements: 1. The straight line joining the object to its image is perpendicular to the surface of the plane mirror.2. A person stands in front of a plane mirror at a distance of (3) meters:

a. The distance between the person and his image = 6 meters.b. If the mirror moves a distance of one meter in the direction of the person, so the distance of the image from the first image is 1 meter(s).

2. Write the scientific term for each of the following: 1. A mirror which gives virtual, erect and equal in size image for an object. Plane mirror.

3. Give reasons for: 1. The image formed by a plane mirror is virtual.Because it cannot be received on a screen.

2. The word AMBULANCE is written in a converted way on the ambulance car.In order to appear in the mirrors of the cars in front of the ambulance car written in a correct way and can be read by the drivers.

3. When you look at a plane mirror, you find that you hold the pen by the left hand which is inverse the real position.Most of people can’t write by a correct way, while they are seeing their writings through a plane mirror.Because the images formed by the plane mirror are laterally inverted (reversed).

Lesson One: Mirrors.Sheet (4): Part Three: Plane Mirror.

1. Study the following figures and answer the following questions: 1. Complete: The opposite figure after redrawing it in your answer sheet. Trace the incident light ray on two mirrors (Y) and (X).

2. Each of the following figures shows the incident light ray on the surface of plane mirror. Answer the following:

a. Calculate the value of angle of incidence in figure (A). 60 o . b. What is the value of angle of reflection in figure (B)? 0 o .




3. A candle was put in front of a plane mirror at a distance of 4cm. An image was formed on the mirror as it is shown in the opposite figure:Which shape (A), (B) or (C) represents the image of the candle? What are its properties? (4 only).

4. Shape (B). The image is upright, virtual, laterally inverted and equal to the object.

5. A triangle ABC is put in front of a plane mirror (as) in the opposite figure:a. Redraw the figure in your notebook and show the formed image for the triangle.b. Mention two properties only of the formed image.

The image is upright and laterally inverted.

6. Noha stands at a distance of 3.5m from a plane mirror and there is a barrier behind her at a distance of 1m.What is the distance between Noha and the image of the barrier in the mirror?Distance between Noha and the mirror = 2×3.5=7m.The distance between Noha and the image of the barrier = 7+1=8m.




Lesson One: Mirrors.Sheet (5): Part Five: Some Concepts Related to Concave Mirror.

1. Complete the following statements: 1. The concave mirror is a part of a sphere, its inner surface is the reflecting surface and it converges light rays after reflection.

2. Convex mirror diverges the light rays and its reflecting surface is the outer surface of a sphere.3. The point that is in the middle of the reflecting surface of the spherical mirror is called the pole of the mirror.4. The center of mirror curvature lies in front of the concave mirror and behind the convex mirror.5. The straight line that passes by the pole of the mirror and its center of curvature is principal axis of the mirror.6. Focus is mid-point between pole of the mirror and center of mirror curvature.7. The distance between the focus of the concave mirror and its pole is called the focal length of the mirror.

8. Focal length = r2 .

9. Real image can be received on a screen, while virtual image cannot.2. Write the scientific term for each of the following:

1. A point that is in the middle of the reflective surface of the spherical mirror.The mid-point on the reflecting surface of the mirror. The pole of the mirror.

2. The distance between the center of curvature of the mirror and any point on its surface.Twice the focal length of a spherical mirror.The radius of the sphere that the mirror is a part of it. Radius of mirror curvature.

3. The straight line that passes by the pole of a spherical mirror and its center of curvature. The principal axis of the mirror.

4. The point at which the incident light ray reflects on itself.The center of the sphere that the mirror is considered a part of it. Center of mirror curvature.

5. The straight line that passes by the center of curvature of the mirror and any point on its surface except the pole of the mirror. The secondary axis of the mirror.

6. The point of collection of the parallel rays after being reflected from the mirror.A point of collection of parallel rays which are parallel to the principal axis of the concave mirror. Focus of the mirror.

7. The distance between the pole of the mirror and its focus. Focal length of the mirror.8. Light rays produced by a distant (far) object. Parallel rays.9. The image that can be received on the screen. Real image.10. The image that cannot be received on the screen.The image formed by the convex mirror or by the concave mirror and can’t be received on the screen.

Virtual image.3. Give reasons for:

1. The spherical mirror has only one principal axis and uncountable number of secondary axes.It has one principal axis, because it has one center of curvature and one pole, while it has uncountable number of secondary axes,, because any straight line passes by its center of curvature except the principal axis is considered a secondary axis.

2. The focal length of a spherical mirror can be determined by knowing its radius of curvature.

Because focal length (f ) = 12×radius of curvature (r)

3. The real image can be received on a screen, while virtual image cannot.Because real image is formed in front of the mirror from intersection of the reflected light rays, while virtual image is formed behind the mirror from the intersection of the extensions of the reflected rays.




Lesson One: Mirrors.Sheet (6): Part Six: Rules to Draw Images Formed by Concave Mirror.

1. Complete the following statements: 1. The light rays produced by a far object are parallel rays.2. The radius of the concave mirror equals twice of its focal length.3. The incident ray that passes through the center of curvature of a spherical mirror reflects back on itself while the light ray which passes through its focus reflects parallel to the principal axis.

4. The incident light rays parallel to the principal axis of the concave mirror, they will reflect passing through the focus.

2. Choose the correct answer: 1. Which of these figures represents the correct path of the reflected rays from a concave mirror? …………………

2. If an incident light ray falls parallel to the principal axis of a concave mirror as in the figure. Which of the reflected rays represent the correct path? …………………

a. 1.b. 2.c. 3. d. 4.

3. What happens when: 1. A light ray is incident on a concave mirror passing through its focus.It reflects parallel to the principal axis.

2. Incidence of a light ray parallel to the principal axis of a concave mirror.It reflects passing through the focus.

4. Show by drawing the path and the direction of rays in the following cases: 1. Parallel ray is incident on a concave mirror and parallel to the principal axis.




Lesson One: Mirrors.Sheet (7): Part Seven: Steps to Draw Images Formed by Concave Mirror.

1. Complete the following figures:

2. Show by drawing the path of the reflected rays in the following cases:




Lesson One: Mirrors.Sheet (8): Part Eight: Formation of Images by Concave Mirror.

1. Complete the following statements: 1. The distance between the image and the concave mirror depends on the distance between the object and concave mirror.

2. To obtain a real, inverted and magnified image for a body, we must put it in front of a concave mirror between the focus and the center of curvature.

3. When a body lies in front of a concave mirror at a distance more than the double of its focal length, a real, smaller and inverted image is formed.

4. A virtual, erect and enlarged image can be formed by a concave mirror.5. If a body is put in front of a concave mirror at a distance of 20cm., and its image is formed in front of the mirror at a distance of 30cm., so the focal length of this mirror is 10 cm.

6. If an object of 8cm. length is put at a distance of 10cm. from a concave mirror its focal length is 5cm., so the length of the formed image is 8 cm.

2. Give reasons for: 1. To obtain a suitable image for you, you should stand at a distance less than the focal length of a concave mirror.Because the formed image will be virtual, upright and magnified.

Lesson One: Mirrors.Sheet (9): Part Eight: Formation of Images by Concave Mirror.

1. Show by drawing the path and the direction of rays in the following cases: 1. An object in front of a concave mirror at a distance equals to the double of its focal length.The formation of the image of a body at the center of curvature of concave mirror.(Determine the properties of the formed image).Properties of the formed image: real, inverted and equal to the object.

2. An object in front of a concave mirror at a distance less than its focal length.(Determine the properties of the formed image).

Properties of the formed image: virtual, erect and magnified.

3. An object in front of a concave mirror at a distance of 7cm, knowing that its focal length is 5cm.The formation of the image of a body between the center of curvature of a concave mirror and its focus.(Determine the properties of the formed image).Properties of the formed image: real, inverted and magnified.

2. Problems: 1. A concave mirror its focal length is 8 cm. If a body is placed at 6 cm from the pole of the mirror show by drawing the image formed and mention the properties of this image.

Properties of the formed image:Virtual – Erect – Magnified.




3. An object is put at a distance 20cm. from a mirror. The image is formed on a screen and has a length equal to the object.

1. What is the type of the mirror?Concave mirror.

2. Calculate the focal length of the mirror.10 cm.

3. Draw the path of rays that show the formation of this image.

4. If you looked at a mirror and found your image is erect and magnified. What is the kind of the mirror and at which distance you exist?Concave mirror, and I am between the focus and the pole.

5. Draw the opposite figure in your answer notebook and by drawing two light rays only determine the position and characteristics of the formed image.

a. The image is formed between the focus and center of curvature. The image is real, inverted and small.

b. The image is formed at a distance greater than the radius. The image is real, inverted and magnified.

c. The image is at the center of curvature. The properties of the image: real, inverted and equal to the object.

Lesson One: Mirrors.Sheet (10): Part Nine: Uses of Concave Mirror.

1. Give reasons for: 1. Concave mirror is used in solar ovens.Concave mirror is used in cooking by using solar energy.To collect a large amount of the solar energy in the focus of the mirror for cooking food.

2. Concave mirror is used to generate high heat energy.Because the concave mirror collects the reflected light rays falling on it in one point (focus) generating high heat energy.

2. What happens when: 1. A plane mirror is put at the left side of the driver of the car instead of a convex mirror.An equal image for the way is formed, so the driver doesn’t see the way.




Lesson One: Mirrors.Sheet (11): Part Ten: Convex Mirror.

1. Complete the following statements: 1. The plane mirror always forms equal virtual image, while the convex mirror forms small virtual image.2. Convex mirror is used as side-view for the driver to form an erect and smaller image for the way behind the car.3. A convex mirror has a focal length of 20cm., then the radius of curvature of its spherical surface equals 40cm.

2. Write the scientific term for each of the following: 1. A mirror whose reflecting surface is the outer surface of a sphere and diverges the light rays. Convex mirror.2. The spherical mirror which forms virtual, upright and small image for the object, wherever its position in front of the mirror. Convex mirror.

3. Give reasons for: 1. The convex mirror is called diverging mirror, while the concave mirror is called converging mirror.Because convex mirror diverges light rays after reflection, while concave mirror converges light rays after reflection.

Lesson One: Mirrors.Sheet (12): Part Eleven: Image Formed by Convex Mirror.

1. Give reasons for: 1. The focus is virtual by the convex mirror.Because this focus is produced due to the collection of the extensions of the reflected rays.

2. The image formed by a convex mirror is always virtual.Because it is formed behind the mirror from the intersection of the extensions of the reflected light rays and can’t be received on a screen.

3. A convex mirror is put at the left side of the driver of the car.Because it forms an erect and smaller image for the way behind the car.

2. Compare between: 1. Concave mirror and convex mirror.

Concave mirror (Converging mirror) Convex mirror (Diverging mirror)A mirror, its reflecting surface is a part of the inner surface of the sphere.

A mirror, its reflecting surface is a part of the outer surface of the sphere.

It converges (collects) light rays after reflection. It diverges light rays after reflection.2. The properties of the image formed by a concave mirror and a convex mirror for an object put between the center of curvature and the focus.

The properties of the image formed by a concave mirror

The properties of the image formed by a convex mirror

Real, inverted and magnified. Smaller than the object, upright and virtual.3. What are the properties of the image formed by the convex mirror wherever the position of the object? Mention one use only for this mirror.The image is smaller than the object, upright and virtual.It is used as side-view mirror on the passenger's side of a car.




Lesson Two: Lenses.Sheet (1): Part Two: Convex and Concave Lenses.

1. Complete the following statements: 1. The focus of the lens is the point of collection of the parallel light rays after refraction from the lens.2. The focal length of the convex lens equals the distance between the focus and optical center of the lens.3. The radius of the convex lens = double of its focal length.4. The focal length of a convex lens equals 10 cm., the radius of surface is 20 cm.

2. Write the scientific term for each of the following statements: 1. The converging lens. The convex lens.2. The optical piece which is thick at the tips and thin at the middle and diverges the light rays falling on it.The concave lens.

3. The optical piece that is thick at the center and less thickness at the tips. The convex lens.4. The point inside the lens on the principal axis in the mid distance between its faces. The optical center of the lens.

5. Half the diameter of the sphere, where the face of the lens is a part of it. The radius of curvature of the face of the lens (r).

6. The straight line joining between the two centers of curvature of the lens and passing through the optical center.The principal axis.

7. The mid-point on the surface of the lens, through which light rays pass without refraction. The optical center of the lens.

8. Any line passes by the optical center of the lens except the principal axis. The secondary axis.9. The point of collection of the refracted light rays which fall parallel to each other and parallel to the principal axis. The principal focus of the lens.

10. The distance between the focus and the optical center. The focal length of the lens.3. Give reasons for:

1. The convex lens is called converging lens, while the concave lens is called diverging lens.Because the convex lens refracts the rays towards the principal axis, while the concave lens refracts the rays away from the principal axis.

2. The focal length of the thick convex lens is less than that of the thin convex lens.Because the radius of the thin lens is bigger than that of the thick lens.

3. The lens has two centers of curvature.Because it has two circular surfaces, each surface has center.

4. The collective lens has two foci, but the collective mirror has one focus.Because convex lens has two circular surfaces, but the concave mirror has one circular surface.

Lesson Two: Lenses.Sheet (2): Part Two: Convex and Concave Lenses.

1. Define each of the following: 1. The convex lens.It is thick at the center and less thickness at the tips.It collects light rays falling on it, so it is called converging lens.

2. The center of curvature of the lens face.It is the center of the sphere, where this face is a part of it.

3. The principal axis of the lens.It is the straight line between the two centers of the lens passing by the optical center of the lens.

4. The focus of the lens.It is the point of collection of the parallel light rays after refraction from the lens.face is a part

2. Study the following figures, then answer the questions: 1. In the following figures, label the points (X), (Y) and (Z).




a. X: The principal axis. b. Y: The optical center of the lens. c. Z: The focal length of the lens.

Lesson Two: Lenses.Sheet (3): Part Three: Rules to Draw the Images Formed by Convex Lens.

1. Complete the following statements: 1. The incident ray that passes through the optical center of the convex lens, it passes in a straight line without refraction.

2. What happens when: 1. A light ray passes through the optical center of the lens.It passes through the lens without refraction.

2. A light ray is incident parallel to the principal axis of the convex lens.It refracts passing through the focus.


4. Show by drawing, the path and directions of rays in the following cases: 1. A light ray is incident on a convex lens passing through its focus.

Show an activity by drawing the determination of the focal length of the convex lens.Materials: Convex lens – Source of light – Screen.

Steps ObservationFix the lens where the distant light source is facing one of its faces.

Parallel rays fall on the lens.

Move the screen closer and farther from the other side of the lens until you get the lit point.

The rays after being refracted are collected in lit point called the focus of the lens that can be received on the




screen.Measure the distance between the lit point and the optical center of the lens.

This distance = ½ radius of curvature of the face of the lens and it is called the focal length of the lens.

Conclusion: Focal length = ½ the radius of curvature.

1. Examine the following figures, then:

a. Mention the figure number which is incorrect “Give reason”.Figure no. (3) Because it doesn’t pass through the focus and exits the lens parallel to the principal axis.

b. Redraw the figure after correction in your sheet paper.




Lesson Two: Lenses.Sheet (4): Part Four: Images Formed by Convex Lens.

1. Complete the following statements: 1. An object with 6cm. height is placed at a distance 10cm., from a convex lens, its focal length is 5cm., therefore the height of the formed image is 6cm.

2. If an object is put in front of a convex lens at a distance of 30cm. and its image is formed behind the lens at a distance of 30cm, so the focal length of the lens is 15cm.

3. When an object is placed between the focus and the center of curvature, the formed image is real, inverted and magnified.

4. No image is formed for the object, when it is at the focus of the convex lens.5. A convex lens has a focal length of 50cm. An object is placed at a distance of 50cm. from the lens the image is formed at infinity.

6. When an object is placed at the focus of a convex lens, the rays penetrate as parallel.2. Complete the following figures:

3. Write the scientific term for each of the following statements: 1. The image which is formed due to the collection of the refracted rays and can be received on a screen.Real image.




Lesson Two: Lenses.Sheet (5): Part Four: Images Formed by Convex Lens.

1. Give reasons for: 1. The object that is placed at the focus of a convex lens has not an image.Because the penetrating ray from a lens doesn’t meet and pass through a parallel way at infinity.

2. The convex lens is used for burning paper with sun rays.Because convex lens collects and directs sunlight in a point (focus) which is burned.

3. The image formed by the convex lens can be received on the screen.Because it is a real image formed as a result of the intersection of the refracted rays.

2. What happens when: 1. You move a screen closed and farther from a convex lens, when it’s other side is facing to a light source.The rays after being refracted collect in one lit point is called the focus of the lens that can be received on the screen.

2. Concentrating sunlight by a magnifying lens on a piece of paper.The piece of paper is burned.

3. What happens when: 1. An object is put at the focus of a convex lens.No image is formed.

2. You want to see a virtual, erect and magnified image of an object through a convex lens.You must put the object before the focus.

4. Show by drawing, the path and directions of rays in the following cases: 1. A lighted body is put in front of a convex lens at a distance of 30cm, knowing that its focal length is 15cm.

2. A lighted body is put before the focus of the convex lens.

3. An object is put at the focus of the convex lens.




4. The formation of the image that is equal to the object by a convex lens.

5. Mention the position and properties of the image formed of an object by means of a convex lens in each of the following cases:

1. The object is at a distance greater than the double of the focal length.Position: Between the focus and the center of curvature.Properties: Real, inverted and small.

2. The object is at a distance equal to twice the focal length.Position: At the center of curvature.Properties: Real, inverted and equal to the object.

3. The object is at a distance greater than the focal length and smaller than twice the focal length.Position: After the center of curvature.Properties: Real, inverted and magnified.

4. The object is put at a distance less than the focal length.Position: After the position of the object in the same side.Properties: Virtual, erect and magnified.

6. A body of length 4cm. at a distance of 6cm. from convex lens, its focal length is 3cm.1. Draw a diagram to show the path of the rays falling on the lens and the refracted ones from it.

2. Mention the properties of the formed image and showing the length of the image and the radius of the lens.Properties of the image: Real – Inverted – Its length is 4 cm.Radius of the lens = 6 cm.

Lesson Two: Lenses.Sheet (6): Part Five: Images Formed by Concave Lens.

1. Complete the following statements: 1. The image formed by the concave lens is always virtual, erect and small.2. It is impossible to obtain real image by using Concave lens.3. Telescope is used for formation enlarged images for the heavenly bodies, while microscope is used for formation magnified images of the tiny bodies.





3. Write the scientific term for each of the following statements: 1. The point of collection of the extensions of refracted rays by a concave lens. The virtual focus.

4. Give reasons for: 1. The image formed by the concave lens cannot be received on the screen.Because it is a virtual image formed as a result of the intersection of the extensions of the refracted rays.

2. It is impossible to obtain a real image by using a concave lens.Because the real image is formed as a result of the intersection of the refracted rays.

5. Show by drawing, the path and directions of rays in the following cases: 1. A virtual, erect and small image is formed for an object put in front of a concave lens.

6.One of the students approaches a lens to one of his eyes and sees through it, he observes that the image of object seems erect. After the lens becomes far to a certain distance from one of his eyes he observed that the image of object seems inverted. The student concludes that the lens must convex:

2. Is the conclusion of the student correct or incorrect? Correct.3. Explain your answer?Because when the lens is near to eye (at a distance less than the focal length) all images formed for objects are erect, while when the lens is far from the eye (at a distance greater than the focal length) all the images formed for objects are inverted.

6. Mention some applications on lenses.1. In designing optical devices (Telescope - Microscope). 2. In making of medical glasses to treat the vision defects.

7. Study the following figures, then answer the questions: 1. The opposite figure to a concave lens has a focal length equals 3cm. An object is placed at a distance of 4cm. from the lens; determine the position of the formed image and its properties by drawing the light rays.

Position: In front of the object at the same side.Properties: Virtual – Erect – Small.




Lesson Two: Lenses.Sheet (7): Part Six: Vision and Vision Defects.

1. Complete the following statements: 1. The concave lens is thin at its center and thick at the tips. It diverges the light rays falling on it and is used in treatment of short-sightedness

2. The normal person can see clearly the near objects at a distance not less than 25cm. and far objects at a distance up to 6m.

3. From the most important vision defects are short-sightedness and long-sightedness.4. The short-sighted person needs a medical eye glasses with concave lenses.5. The vision defect which is due to a shortness in the radius of the eye sphere (ball) is called long-sightedness.6. In short-sightedness, the images of the far objects are formed in front of retina, while in long-sightedness, the images of the near objects are formed behind the retina.

7. A long-sighted person needs a medical eye glasses with convex lens.2. Write the scientific term for each of the following statements:

1. Seeing the near objects clearly and seeing he far objects distorted. Short-sightedness.2. A disease in the eye due to weakness of the eye lens or shortness of the eyeball diameter.A vision defect results due to the formation of the image behind the retina of the eye. Long-sightedness.

3. The lens that corrects the short-sightedness. The concave lens.4. The lens that corrects the long-sightedness. The convex lens.

3. Give reasons for: 1. Short-sighted sees the far objects distorted.Because the images of these objects do not fall on the retina of the eye but in front of it.

2. Some persons have short-sight.Due to:

a. The increase in the eyeball diameter. b. The increase in convexity of the eye lens surface.

3. In short-sight, the retina is far from the eye lens.Due to the increase in the eyeball diameter.

4. Concave lens is used to treat short-sighted person.To diverge the rays coming from a far object, so the image is formed on the retina.

5. Some persons have long-sight.Due to:

a. The decrease of the eyeball diameter. b. The decrease of convexity of eye surface.

6. Long-sighted person cannot see the close objects clearly.Because the image of the close objects don not fall on the eye retina but behind it.

7. The retina is close to the eye lens in the long-sighted person.Due to the decrease of the eyeball diameter.

8. The near objects are collected behind the eye retina in long-sightedness.Due to the decrease of convexity of eye surface which results more focal length, so the rays coming from the near object are collected in a point behind the eye retina.

9. The long-sightedness is treated by using a convex lens.Because the convex lens collects the rays, so the image of the object is formed on the retina.




Lesson Two: Lenses.Sheet (8): Part Six: Vision and Vision Defects.

1. What happens when: 1. The eye lens is too convex.He can see near objects clearly but far objects seem distorted.

2. The eye lens surface in man is less convex.He can see far objects only clearly but close objects are not seen clearly.

3. The diameter becomes longer than a certain length.This causes the retina to be far from the eye lens.

4. The diameter of the eye ball becomes shorter than a certain length.It causes long-sightedness.

5. The eyeball diameter decreases.This causes the shortness of the radius of the eye sphere, thus the retina is close to the eye lens and this causes long-sight.

2. Compare between: 1. Long-sight and short-sight. (Concerning: The type of lens used in treatment of each one – the cause of each one).

P.O.C Long-sightedness Short-sightednessThe correction: By using a convex lens. By using a concave lens.Causes: - The diameter of the eyeball is too short.

- The lens is thin.- The diameter of the eyeball is too large.- The lens is too convex.

3. Mention briefly the reasons of short-sightedness and how to correct it.Reasons:

1. The increase in the eyeball diameter. 2. The increase in convexity of the eye lens surface. Corrections: By using a concave lens which disperses the rays coming from a far object, so the image is formed on the retina.

4. Mention briefly the reasons of long-sightedness and how to treat long-sighted person.Reasons:

1. The decrease in the eyeball diameter. 2. The decrease in convexity of the eye lens surface. Corrections: By using a convex lens which collects the rays, so the image is formed on the retina.

5. Two friends Ahmed and Ali were reading at the school library. Ahmed noticed that his friend was reading the only books which are far from his eyes:

1. What’s wrong with Ali?Long-sight.

2. How can he solve his problem?By using a convex lens

3. Correction (treatment) of short-sightedness.4. Correction (treatment) of long-sightedness.




Unit Three: The Universe and the Solar System.Lesson One: The Universe.

Sheet (1): Part One: Universe and Solar System.1. Complete the following statements:

1. The universe is the space which contains all the galaxies, stars planets, moons and living organisms.2. Galaxies are groups of stars that rotate together in space by the effect of gravitational force.3. Each galaxy has a distinctive shape according to harmony and order of the groups of stars in universe.4. The stars move in fixed orbits around the center of the galaxy.5. The galaxy that our earth planet belongs to is called Milky Way galaxy.6. The building unit of the universe is Galaxy and its number in the universe is about 100,000 million.7. The galaxy is the structural unit of the universe and our galaxy is Milky Way.8. Galaxies gather in clusters including the Milky Way which contains solar system.9. In Milky Way galaxy, the old stars gather in the center while small stars are located in the spiral arms of the galaxy.

10. The solar system is located in one of the spiral arms of Milky Way galaxy.11. The solar system contains a number of planets orbit Sun.12. The distances between stars are measured in light year unit and a equals 9,460,000 million km.

2. Write the scientific term for each of the following statements: 1. The space which contains all the galaxies, stars, planets, moons, living organisms and everything. The universe.

2. The bigger units that form the universe. The galaxy.3. It is the wide and extended space that contains galaxies. The universe.4. It contains all the stars we see at night in the sky. The Milky Way galaxy5. The galaxy that is our solar system belongs to. The Milky Way galaxy6. A galaxy appears in the sky at night as scattered straw or spilled milk. The Milky Way galaxy7. They are located in the spiral arms of the Milky Way galaxy. Small stars.8. The sun and eight planets revolving around it. The solar system.9. It is located in one of the spiral arms of the Milky Way galaxy. The solar system or the Sun.10. The unit that is used to measure the distances between the celestial bodies. Light year.11. The distance that is covered by light in one year. Light year.



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Lesson One: The Universe.Sheet (2): Part One: Universe and Solar System.

1. Give reasons for the following: 1. Our galaxy is called by Milky Way galaxy.Because it appears in the sky at night as a splashing milk or spreading straw.

2. Astronomers don’t measure the distances between stars in kilometers.Because the distances between stars are very large.

2. What is meant by: 1. The universe.It is the space which contains all galaxies, stars, planets, moons, living organisms and everything.

2. Galaxies. They are the bigger units of the universe.3. Light year.It is the distance covered by the light in one year and it equals 9.467 × 10 12 km.

3. Variant questions: 1. Study the opposite figure, then answer:

a. What’s the galaxy which our solar system belongs to?The Milky Way Galaxy.

b. What does point (X) refer to?The Sun.

2. The following figure represents the components of the Universe.Label the figure:

1. The Universe. 2. Galaxies. 3. The Milky Way Galaxy. 4. The Solar System. 5. The Earth Planet.

Lesson One: The Universe.Sheet (3): Part Two: Big Bang Theory.

1. Complete the following statements: 1. The first human perception of cosmogony was in the Stone age, where Myths dominated the human imagination.

2. Ancient Egyptians and Babylonians put a relationship between the eternal universe and the multiple gods controlling it.

3. The scientists believe that the universe originated from a massive explosion called the Big Bag.4. Scientists believe that the matter of the universe was a gaseous ball of high pressure and temperature in a small volume.

5. The Big Bang theory had been developed since1933.6. Within minutes of the Big Bang, the atomic particles merged together producing helium and hydrogen gases.7. Galaxies began to from after 3000 million years after the Big Bang.8. Milky Way galaxy took its disc form after 5000 million years of the Big Bang.9. In 1964, scientists coincidently discovered radio waves coming from the space and they concluded these waves are a type of the echo coming from the Big Bang.

2. Write the scientific term for each of the following statements: 1. The continuous separation between galaxies. Expansion of the universe.2. The expansion of universe and merging of atomic particles forming helium and hydrogen gases.A theory explains the origin of the universe due to a great explosion, followed by continuous expansion and changes since 15000 million years. Big Bang.




3. Give reasons for the following: 1. The continuous expansion of the space.Due to the continuous spacing between galaxies in the universe.

2. Galaxies move away from each other.Because the universe continuous in a state of expansion.

3. The gravity has important role in cosmogony of the universe.Because it helped in gathering of the masses of the matter of the universe.

4. Forming of an atmosphere around some celestial bodies.Due to strong gravity which attracts gases towards the planets forming an atmosphere.

4. What is meant by: 1. Big Bang.It is massive explosion followed by expansion of the universe and merging of atomic particles forming Helium and Hydrogen gases.

Lesson One: The Universe.Sheet (4): Part Three: Future of Universe.

1. Variant questions: 1. Compare between the closed universe theory and opened universe theory.

Open universe theory Closed universe theoryIn this theory scientists see that there is no definite end to the universe.

In this theory scientists see that the universe will stop expanding and will begin to contract until it becomes very compact or very hot preparing for a new Big Bang.

2. What are the results of:The universe stops expanding in the closed universe theory.The universe will begin to contract until it becomes very compact or very hot preparing for a new Big Bang.

Unit Three: The Universe and the Solar System.Lesson Two: The Solar System.Sheet (1): Part One: Gravity.

1. Complete the following statements: 1. The solar system consists of The Sun, eight planets, asteroids, comets and moons.2. The Sun is the center of solar system and it represents 99% of the total mass of the solar system.3. The force of attraction between two objects is directly proportional to the product of their masses and is inversely proportional to the square of the distance between them.

4. If the gravitational force was strong enough, it will attract gases towards the planet or the moon forming an atmosphere around it.

5. As the distance between the planet and the sun increases, the Sun’s gravitational force decreases and its motion becomes slower.

6. The moon rotates around the earth in a fixed orbit and Earth rotates around the Sun once every earthly day.7. Planets and other bodies were originated in the solar system from the matter that remained from the evolution of the Sun.

8. The solar nebula is considered a gaseous cloud of hydrogen and helium gases and dust of iron, rocks and ice.9. When the solar nebula turned into a flat rotating disk, most of dust compressed together forming Mercury, Venus, Earth and Mars, while some of the dust and ice combined with gases forming Jupiter, Saturn, Uranus and Neptune.

2. Write the scientific term for each of the following: 1. The space which contains all the galaxies, stars, planets, moons, living organisms and everything. The Universe.

2. It includes the Sun, planets, moons and other celestial bodies. The solar system3. The biggest star can be seen clearly by people on the Earth. The Sun4. Eight planets rotate around the Sun. Planets of the solar system




5. It is located in one of the spiral arms of the Milky Way galaxy. Solar System6. The force that keeps the continuity of the planets rotation in their orbits around the Sun. Gravitational force7. The force of attraction between two objects is directly proportional to the product of their masses and is inversely proportional to the square of the distance between them. Newton's law of universal gravitation.

Lesson Two: The Solar System.Sheet (2): Part One: Gravity.

1. Correct the underlined words: 1. Nine planets rotate around the sun. Eight2. The solar system contains eight of stars. Planets3. The solar system extends over 4500 million kilometer in space. 12000 million4. When you are inside a lift going downward fast, you feel that you are heavier in weight. Lighter

2. Give reasons for: 1. The Sun is the dominant body in the solar system.Because it represents more than 99% of its total mass.

2. The stability of the Earth rotation in an orbit around the Sun.Due to the central gravitational force of the Sun.

3. The constancy of the planets in their orbits around the Sun.Due to the central gravitational force of the Sun.

3. What would happen … ? 1. If the gravity between the Sun and planets which rotate around is vanished.The planets will leave their orbits and float in a random fashion in space and this leads to the destruction of the Earth.

2. What is meant by … ? 3. The solar system.It is a system includes the sun, eight planets, comets, asteroids and moons.

4. General Gravitational Force.It is the force that keeps the continuity of the planets rotating in their orbits around the Sun.

5. Newton's law of universal gravitational force.The force of attraction between two bodies is directly proportional to the product of their masses and is inversely proportional to the square of the distance between them.

Lesson Two: The Solar System.Sheet (3): Part Two: Nebular Theory.

1. Complete the following statements: 1. Scientist Laplace has been affected by two observations which are presence of nebula in the space and presence of rings of planet Saturn.

2. The scientist who found Nebular theory about the evolution of the solar system is called Laplace.3. In the nebular theory, over the time, the nebula lost its heat gradually, so its size contacted and its revolving speed around itself increased.

4. According to nebular theory, the gaseous rings cooled down and frozen forming the planets, while the flaming mass remained is the Sun.

2. Write the scientific term for each of the following: 1. A flat gaseous round disk that formed the solar system. Nebula2. A theory assumed that the solar system was originally a glowing gaseous sphere revolving around itself.

Nebular theory3. Correct the underlined words:

1. After turning the solar nebula into a flat rotating disk, the dust and ice combined with gases forming Mercury, Saturn, Venus and Neptune. Jupiter – Uranus







4. Give reasons for: 1. The nebula lost its sphere form and became in a form of a flat rotating disk.Due to the effect of the centrifugal force.

5. What would happen … ? 1. When the nebula lost its temperature in Laplace's opinion.Its size contacted and its revolving around itself increased.

6. Mention: 1. The planets that contain mixture of ice and dust in their composition.Jupiter – Saturn – Uranus – Neptune.

7. Variant questions: 1. Explain the evolution of the solar system as the vision of the French scientist Laplace?

a. The Nebula [Glowing gaseous sphere]: Over the time, the nebula lost its heat gradually, so its size contracted and its revolving speed around itself increased.

b. The gaseous rings: 1. Under the effect of centrifugal force. The nebula lost its spherical shape and became in a form of a flat disk.2. Parts got separated from the flat rotating disk to form gaseous rings that also rotate in the same direction in which the nebula rotates.

c. The formation of the solar system: 1. The gaseous rings cooled down and frozen forming the planets of the solar system.2. The flaming mass that remained in the center formed the Sun.

Lesson Two: The Solar System.Sheet (4): Part Three: Crossing Star Theory.

1. Complete the following statements: 1. In the crossing star theory, the explosion of the crossing star led to formation of a gaseous line of a great and also the Sun escaped from the gravity of this star.

2. Write the scientific term for each of the following: 1. A theory assumed that the solar system was originally a huge star (the Sun). The crossing star theory

3. Give reasons for: 1. The escaping of the Sun from the gravity of the huge star in the crossing star theory.Due to the explosion of the expanded part of the Sun.

2. Chamberlain and Moulten theory is called by the crossing star theory.Because it assumes crossing of a huge star near the Sun.

Lesson Two: The Solar System.Sheet (5): Part Four: Modern Theory of the World.

1. Complete the following statements: 1. The scientist who established Nebular theory is Laplace, but the scientist who established the modern theory of the world is Alfred Hale.

2. In Alfred Hale theory, a cloud remained from the explosion, and it subjected to cooling and contraction processes forming the matter of planets.

1. Write the scientific term for each of the following: 1. A theory assumed that the solar system was originally a star rather than the Sun. The modern theory of the world

2. Correct the underlined words: 1. The modern theory for formation of the solar system according to Laplace is due to explosion of a star rotating around the Sun. Alfred Hale.




Lesson Two: The Solar System.Sheet (6): Part Five: Day and Year Difference between Planets.

1. Complete the following statements: 1. The Earth revolves around its axis in a period of time estimated at one earth’s day and it revolves around the Sun in a period of time estimated at one earth’s year.

2. The difference of the length of a day from a planet to another is due to radius of the planet and speed of rotation around its axis.

3. The difference of the length of a year from a planet to another is due to distance between planet and the Sun and speed of rotation around the Sun.

4. The length of the day on Saturn equals 0.43 earth’s day, while on Mercury equals 59 earth’s day.5. The longest day is on Venus planet, whereas the shortest day is one Jupiter planet.6. The shortest year is the year of Mercury planet.7. On Neptune, the year is 0.24 earth’s year, while on Uranus is 84 earth’s year.8. Jupiter rotates around the Sun once every 12 earthly years.

2. Write the scientific term for each of the following: 1. The time of revolving of the earth around its axis. The Earthly day2. The time taken by the planet to complete one rotation around its axis. The day3. The time taken by the planet to complete one rotation around the Sun. The year4. The planet that has the shortest year on its surface. Mercury5. The planet that has the longest year on its surface. Neptune6. The planet that has the shortest day on its surface. Jupiter7. The planet that has the longest day on its surface. Venus

3. Correct the underlined words: 1. The difference of a day length from a planet to another is due to the speed of the planet rotation around the Sun.

Its axis2. The complete rotation of Mars planet around the Sun takes a time equals 12 earth's year. 1.9 earth's year3. The time taken by Saturn planet to rotate around the Sun is 89 years. 29 years4. The time of revolving Venus planet around its axis is one Earthly day. Earth

Lesson Two: The Solar System.Sheet (7): Part Five: Day and Year Difference between Planets.

1. Give reasons for: 1. The difference in the length of the year from a planet to another.Due to the difference in:

a. The distance between the planet and the Sun. b. The speed of rotation of the planet around the Sun.

2. The difference in the length of the day from a planet to another.Due to the difference in:

a. The radius of the planet. b. The speed of rotation of the planet around its axis.

3. The longest year is on Neptune.Because it is the farthest planet from the Sun.

4. The shortest year is on Mercury.Because it is the nearest planet to the Sun.

5. The year in Venus is shorter than its day.Because its speed of rotation around the Sun is much greater than its speed of rotation around its axis.

2. What would happen … ? 1. When the distance between a planet and the Sun increases.The length of the year on its surface will increase.




2. Due to the difference in speed of planet rotation around its axis.The difference in the length of the day from a planet to another.

3. Due to the difference in speed of planet rotation around the Sun.The difference in the length of the year from a planet to another.

Lesson Two: The Solar System.Sheet (8): Part Six: Space Suit – Telescope – Spacecraft.

1. Complete the following statements: 1. Solar telescope is used in studying the Sun from the Earth’s surface.

2. Write the scientific term for each of the following: 1. An equipment was launched to the space allows astronomers an opportunity to study the evolution of the universe after the Big Bang.Hubble telescope

3. Give reasons for: 1. Telescopes rotating around the Earth are preferred than that on the Earth's surface.Because:

a. They can see celestial bodies more clearly. b. They can catch rays that don't able to penetrate the Earth's atmosphere.

Unit Four: Reproduction and Species Continuity.Lesson One: Cell Division.

Sheet (1): Part One: Cells and Chromosomes.1. Complete the following statements:

1. Somatic cells and reproductive cells are two types of cells in the bodies of multicellular living organisms.2. The hereditary material in the nucleus of the cell which consists of a number of chromosomes.3. The chromosome chemically consists of nuclear acid called nucleic acid [DNA] and protein.4. The nuclear acid is symbolized by DNA which carries genes of the living organism.5. The chromosome consists of two chromatids connected together at point known as centromere.6. The number of chromosomes is fixed in the individuals of the same species, while it is different from a species to another.

2. Write the scientific term for each of the following: 1. They are thread like bodies that have the main role in cell division. Chromosomes.2. The part in the cell which is responsible for cellular division. Cell nucleus.3. It consists of two chromatids connected together at centromere. Chromosome.4. The point of connection of two chromatids together. Centromere.5. The nuclear acid that carries the genetic traits. DNA.

3. Give reasons for each of the following sentences: 1. Chromosome is considered the genetic material for the cell.Because it contains the genes that carry the genetic traits of the living organism.




Lesson One: Cell Division.Sheet (2): Part Two: Stages of Mitotic Division.

1. Complete the following statements: 1. There are two types of cell division which are mitotic division and meiotic division.2. Before starting cell division, the cell passes through a phase called interphase in which the amount of genetic material [DNA] duplicates.

3. During prophase, chromatin reticulum intensifies and chromosomes appear in the form of thin strings.4. At the end of prophase of mitosis, the nucleolus and nuclear membrane disappear.5. In the animal cell, the spindle fibers are formed by the centrosome, while in plant cell, the spindle fibers are formed from condensing the cytoplasm at the cell poles.

6. During metaphase, chromosomes are directed to the equator of the cell.7. The spindle fibers are formed during the cell division in prophase phase and disappear in telophase phase.8. During anaphase of mitosis, the centromere splits lengthwise into two halves and the chromatids of each chromosome are separated from each other.

9. A nuclear membrane is formed at each pole of the cell surrounding by the chromosomes in the telophase phase of the cell division.

2. Choose from column (B) what suits in column (A):

(A) (B)1. The spindle fibers shrink and two identical groups of chromosomes are formed at each pole of the cell in.

a. Telophase.

2. Duplication of genetic material occurs in b. Prophase.3. The nucleolus and nuclear membrane disappear in c. Interphase.4. The chromosomes are arranged along the cell equator in

d. Metaphase.

e. Anaphase.1. e. 2.c 3.b 4.d

3. Write the scientific term for each of the following: 1. A cell division that occurs in the somatic cells and results in the growth of the living organism.A kind of cell division which is responsible for compensation of the damaged cells. Mitosis.

2. A phase in which some important vital processes occur which prepare the cell or division and the genetic material in the cell is doubled. Interphase.

3. A phase in which the chromosomes migrate towards the cell equator where each chromosome is connected with one of the spindle fibers at the centromere.The phase in which the chromosomes are arranged at the middle of the cell during its division. Metaphase.

4. The phase of mitotic division in which the nucleolus disappears. Prophase.5. Fibers extend between the two poles of the cell in prophase. Spindle fibers.





1. Give reasons for each of the following sentences: 1. Cellular division begins with interphase. To prepare the cell for division through some important biological processes, where the amount of genetic material duplicates.

2. Duplicating the genetic material in interphase for cell division.To each of the cells resulting from division obtains a complete copy of genetic material.

3. The mitotic cell division is responsible for the growth of living organisms.Because it duplicates and increases the number of body cells.

4. The difference in the way of formation of spindle fibers in plant cell than in animal cell.Because in the plant cell, the spindle fibers are formed condensing the cytoplasm at the cell poles, while in the animal cell, they are formed by the centromere.

5. Shrinking of spindle fibers during the anaphase of mitosis division.To form identical groups of chromosomes.

6. The changes that occur in telophase of mitotic division are called adverse changes.Because they inverse the changes that occur in prophase.

7. The damaged nerve cells can't be compensated.Because nerve cells don't divide at all.





1. Show by drawing each of the following and write down the labels as possible: 1. The phases of mitotic cell division.

2. Metaphase in mitotic division.

2. Variant questions: 1. A cell has (2N) chromosomes in its nucleus and it is divided meiotically. How many cells are produced from such division and how many chromosomes in its nucleus?4 cells – (N) chromosomes.

2. What’s the scientific base of liver transplantation?Mitosis division.

3. From the opposite figure, answer the questions:a. Which phase does the figure represent?Interphase.

b. Mention what happens in this phase (in mitosis)?Duplication of DNA.

4. Examine the opposite figures, then answer the following questions:a. Identify phases (A) & (B).

1. (A) Anaphase. 2. (B) Metaphase.

b. What is the difference between the two phases?In Metaphase chromosomes are arranged along the cell equator, while in Anaphase the chromatids separate from each other.

c. What are the changes which occur in stage (A)?1. The centromere of each chromosome splits lengthwise into two halves, so the chromatids separate from each other.




2. Spindle fibers begin to shrink and two identical groups of chromosomes (each contains single chromatid) are formed.

3. Each group of chromosomes migrates towards one of the cell's poles. d. Write the labels on fig. (B).

1. Centromere. 2. Spindle fibers. 3. Chromosome.

5. Look at the opposite diagram it is for a phase in Mitotic division:a. What is the name of this phase?Metaphase.

b. What does the centrosome do during Mitotic division?By centrosome, the spindle fibers are formed.

c. What happens to the two chromatids of each chromosome during anaphase?During Anaphase, the centrosome of each chromosome splits lengthwise into two halves, so the chromatids separate from each other.

d. In which cells does Mitotic division occur?Somatic cells.

6. Name only these stages of mitosis:

a. (1) Telophase. b. (2) Metaphase. c. (3) Prophase. d. (4) Anaphase.

Lesson One: Cell Division.Sheet (5): Part Three: Meiosis (Meiotic Division).

1. Complete the following statements: 1. Meiotic division aims to form gametes.2. The somatic cells divide by mitosis cell division, while the reproductive cells divide by meiosis cell division.3. In human, meiosis occurs in testis to produce sperms, while it occurs in ovary to produce ova.4. In plants, male gametes are called pollen grains, while female gametes are called eggs.5. Meiosis takes place in two stages which are first meiotic division and second meiotic division.6. In flowering plants, the pollen grains are formed inside the anthers, while the ovules are produced inside the ovary.

2. Write the scientific term for each of the following: 1. Cellular division which leads to the formation of gametes. Meiosis.




2. The cells resulting from meiotic division and have half number of chromosomes (N) of the original cell.Gametes.

3. Give reasons for each of the following sentences: 1. Meiotic division is called by reduction division.Because the produced cells contain half the number of chromosomes of the original cell.

2. The gametes are often (N), while somatic cells are often (2N).Because gametes produced by meiosis division, while somatic cells produced by mitosis division.

3. Mitosis division is important for children than meiosis.Because mitosis division plays an important role in growth of living organisms and compensating the damaged cells.

Lesson One: Cell Division.Sheet (6): Part Four: Stages of First Meiotic Division.

1. Complete the following statements: 1. In prophase I, chromosomes are arranged in homologous pairs, each pair consists of 4 chromatids which are called a tetrad.

2. Each produced cell from meiosis contains half number of chromosomes of the parent cell, so it is called reduction division.

3. The cell which divides by meiotic cell division gives 4 cells, while that divides by mitotic gives 2 cells.4. If the cell of liver in rabbit contains 44 chromosomes, so the male gamete in rabbit contains 22 chromosomes and the female gamete contains 22 chromosomes.

5. Crossing over phenomenon takes place during prophase I of the meiotic division.2. Write the scientific term for each of the following:

1. It occurs at the end of the first prophase of the meiosis, in which the inner parts of chromatids are exchanged.A process in which the inner parts of chromatids of each tetrad are exchanged. Crossing over phenomenon.

3. Give reasons for each of the following sentences: 1. Meiosis is considered the source of genetic variation on which the variation of living organisms depends on.Due to occurring the crossing over phenomenon during it.

2. Crossing over is the source of genetic variation between members of the same species.Because it contributes in genes exchanging between the two homologous chromosome's chromatids and distributing them randomly in the gametes.

Lesson One: Cell Division.Sheet (7): Part Four: Stages of First Meiotic Division.

1. What is meant by: 1. Crossing over phenomenon.It is a phenomenon that takes place at the end of prophase I and in which some parts of the two inner chromatids of each tetrad are exchanged to produce new genetic arrangement.

2. Mention the importance of each of the following: 1. Crossing over phenomenon.It is responsible for the variation of genetic traits among the members of the same species.

3. What is the result of: 1. Each two homologous chromosomes come close to each other to form tetrad.Crossing over phenomenon occurs.

2. The parts of the inner chromatids are exchanged in the first prophase.Crossing over phenomenon occurs.

3. Meiosis division of reproductive cells occurs in human body.It will produce gametes that contain half number of chromosomes.

4. Crossing over occurs at the end of prophase in meiosis.The genetic variation occurs among the members of the same species.







Lesson One: Cell Division.Sheet (8): Part Five: Nanotechnology and Cancer Treatment.

1. Complete the following statements: 1. The Dr. Mustafa El-Said scientist discovered a way to detect the cancer cells by a technic called nanotechnology.

2. Nano-molecules of gold metal are used to detect cells of cancer and laser rays are used to destroy them.2. Write the scientific term for each of the following:

1. A dangerous disease occurs when the body cells are divided continually without controlling. Cancer.3. Give reasons for each of the following sentences:

1. Nanotechnology is called by this name.Because in which, very small molecules are used which their lengths are measured by nanometer unit.

2. Laser is used for treatment of cancer by nanotechnology.Because the Nano-molecules of gold which stuck the surface of cancerous cell absorb the light of laser and convert it into heat which leads to burn and kill the infected cell.

4. Mention the importance of each of the following: 1. Nano-molecules of gold.Treating of cancer.

2. Proteins that are loaded on gold molecules.Attach (adhere) on the cancerous cell to monitor it.

3. Laser in treating cancer by nanotechnology.Burning and killing the infected cell.




Lesson One: Cell Division.Sheet (8): Part Five: Nanotechnology and Cancer Treatment.

5. Complete the following statements: 1. The Dr. Mustafa El-Said scientist discovered a way to detect the cancer cells by a technic called nanotechnology.

2. Nano-molecules of gold metal are used to detect cells of cancer and laser rays are used to destroy them.6. Write the scientific term for each of the following:

3. A dangerous disease occurs when the body cells are divided continually without controlling. Cancer.7. Give reasons for each of the following sentences:

4. Nanotechnology is called by this name.Because in which, very small molecules are used which their lengths are measured by nanometer unit.

5. Laser is used for treatment of cancer by nanotechnology.Because the Nano-molecules of gold which stuck the surface of cancerous cell absorb the light of laser and convert it into heat which leads to burn and kill the infected cell.

8. Mention the importance of each of the following: 6. Nano-molecules of gold.Treating of cancer.

7. Proteins that are loaded on gold molecules.Attach (adhere) on the cancerous cell to monitor it.

8. Laser in treating cancer by nanotechnology.Burning and killing the infected cell

Unit Four: Reproduction and Species Continuity.Lesson Two: Sexual and Asexual Reproduction.

Sheet (1): Part One: Types of Reproduction.1. Complete the following statements:

1. The genetic traits pass from parents to offspring by reproduction process.2. Reproduction is a biological process, where the living organism produces new individuals of the same kind ensuring its continuity.

3. There are two types of reproduction in living organisms which are asexual and sexual4. Asexual reproduction occurs by only one living organism.5. Asexual reproduction is mostly occurring in unicellular living organisms and some multicellular animals and plants.

2. Write the scientific term for each of the following: 1. A biological process, where the living organisms produce new individuals of the same kind to ensure its continuity. Reproduction.

2. The ability of living organisms to produce new individuals. Reproduction.3. A process in which living organisms produce new individuals with genetic traits identical to those of their parents. Asexual reproduction.

4. The type of reproduction that takes place by only one living organism. Asexual reproduction.3. Give reasons for:

1. Reproduction is the way of living organisms to ensure the continuity of their species.Because by reproduction, the living organisms produce new individual of the same kind.

2. All types of a sexual reproduction take place by only one individual.Because asexual reproduction includes mitosis division.

3. Asexual reproduction needs no special systems and structures to occur.Because asexual reproduction occurs by mitosis division through one individual.




Lesson Two: Sexual and Asexual Reproduction.Sheet (2): Part Two: Binary Fission.

1. Complete the following statements: 1. Paramecium reproduces asexually by binary fission.2. In binary fission, the nucleus is divided by mitosis division and the cell splits into two cells.3. Amoeba reproduces asexually by Binary fission.

2. Write the scientific term for each of the following: 1. A type of a sexual reproduction occurs in unicellular living organisms.The nucleus divides mitotically (mitosis) and then the cell which represents the body of the unicellular organism splits into two cells. Binary fission.

3.Give reasons for:

2. Binary fission is considered as mitotic division.Because two identical cells are produced each is identical to the original cell.

Lesson Two: Sexual and Asexual Reproduction.Sheet (3): Part Three: Budding.

1. Complete the following statements: 1. Asexual reproduction takes place by budding in yeast fungus and by binary fission in bacteria.2. In reproduction by budding, one of the two produced nuclei remain in the parent cell and the other migrates to the bud.

3. When the buds remain connected to the parent cell, a colony is formed.4. The yeast fungus reproduces by budding which is considered as a type of asexual reproduction.

2. Write the scientific term for each of the following: 1. The type of asexual reproduction in which a colony may be formed. Budding.2. The type of a sexual reproduction occurs in yeast fungus and sponges. Budding.

3. Give reasons for: 1. A colony may be formed through reproduction by budding.Because the buds remain connected to the parental cells.

4. Variant questions: 1. Show by drawing each of the following and write down the labels as possible:

a. The reproduction in the yeast fungus.

2. Rearrange the following figures to explain the process of asexual reproduction in amoeba and yeast fungus, then mention the type of reproduction in each organism:


Amoeba (a) Yeast fungus (b)(2) (4)(3) (2)(1) (3)(4) (1)

Reproduces by Binary Fission Reproduces by Budding



Lesson Two: Sexual and Asexual Reproduction.Sheet (4): Part Four: Regeneration.

1. Complete the following statements: 1. From the examples of living organisms which reproduce by regeneration is starfish.2. Starfish reproduces asexually by Regeneration.3. If the chromosome number in the cells of starfish is (2N), the number of chromosomes in the produced cells is 2N.

2. Write the scientific term for each of the following: 3. T Give reasons for:

1. Starfish continues alive even a part of its body is cut.Because this part contains a part of the central disc of the animal.

2. The number of chromosomes in cells resulted by regeneration is (2N) as in the parental cell.Because regeneration is asexual reproduction that occurs by mitosis division.

Lesson Two: Sexual and Asexual Reproduction.Sheet (5): Part Five: Sporogony [Spore Propagation].

1. Complete the following statements: 1. In bread mould fungus, sporangia have great numbers of spores which are released on rupturing of their wall.2. Bread mould fungus reproduces asexually by Spore propagation.3. Starfish reproduces asexually by regeneration, while bread mould fungus reproduces by spore propagation.4. Binary fission, budding, regeneration, spore propagation and vegetative reproduction are the types of asexual reproduction.

2. Write the scientific term for each of the following: 1. Structures that are formed inside the sporangia. Spores.2. The most common asexual reproduction in fungi and algae. Sporogony.

3. Give reasons for: 1. Spore propagation is a type of asexual reproduction which is common in some fungi such as bread mould and mushroom.Because it is asexual reproduction occurs by spores present in the sporangia.

4. Mention the importance of: 1. Spores.They grow in the suitable medium producing new individuals identical to the parental individual.

5. What is the relation between the genetic structure of offspring and parents with giving the reasons in each of the following cases:

1. Spore propagation in yeast fungus.The cells are identical and similar to the parental cells. Because they are resulted by mitotic division.

Lesson Two: Sexual and Asexual Reproduction.Sheet (6): part (6): Vegetative reproduction

1. Complete the following statements: 1. Vegetative reproduction in plants needs no seeds, but these plants reproduce by vegetative organs.2. Vegetative reproduction is a type of asexual reproduction in which plants reproduce by using vegetative organs except seeds.

2. Write the scientific term for each of the following: 1. Asexual reproduction by using vegetative organs except seeds. Vegetative propagation.

3. Give reasons for:




1. Vegetative reproduction is called by this name.Because it occurs without need of seeds but by the plants vegetative organs.

4. What is meant by each of the following: 1. Vegetative reproduction.Producing new individuals identical to the parental individual without need of seeds.

Lesson Two: Sexual and Asexual Reproduction.Sheet (7): Part Seven: Sexual Reproduction.

1. Complete the following statements: 1. Sexual reproduction depends on two main processes that are gametes formation and fertilization.2. Zygote results from the combination of a male gamete and a female gamete which contains the same number of chromosomes of the living organism.

3. There are two types of reproduction which are sexual reproduction and asexual reproduction.4. Sexual reproduction depends on two processes which are Gametes formation and fertilization.5. During asexual reproduction, the number of parents is one, while it is two during sexual reproduction.6. The new individuals produced by asexual reproduction have traits identical to their parents. Write the scientific term for each of the following:

7. The type of reproduction in which no genetic variation takes place. Asexual reproduction.8. A process by which the living organism produces individuals with traits differ from parents. Sexual reproduction.

9. The type of reproduction that occurs in higher multicellular organisms. Sexual reproduction.10. A process in which the fusion takes place between a male gamete and a female gamete to form a zygote.

Fertilization.11. The type of reproduction which is considered as a source of genetic variation. Sexual reproduction.12. The structure that is resulted from the combination of a male gamete and a female gamete.Zygote.13. It has genetic material from both parents and during growth gives new individual carries the traits of both parents. Zygote.

2. Give reasons for: 1. Sexual reproduction occurs in most higher living organisms of plants and animals.Because it depends on two main processes, gametes formation and fertilization.

2. The zygote has the same number of chromosomes of cells of parental organism.The individuals resulted from the sexual reproduction are not similar to their parents.Because it is produced from the combination of a male gamete and female gamete, each of them contains half number of chromosomes of the parental organism.

3. The number of chromosomes is constant in the same species which reproduce.Because each of male gamete and female gamete contains half number of chromosomes (N), by combination, a zygote containing the whole number of chromosomes (2N) is formed.

4. Sexual reproduction is a source of the genetic variation between individuals.Because the offspring resulting from sexual reproduction combines the genetic traits from two sources.

Lesson Two: Sexual and Asexual Reproduction.Sheet (8): Part Seven: Sexual Reproduction.

1. What is meant by each of the following: 1. Fertilization.It is the combination of a male gamete and a female gamete to form a zygote which contains the normal number of chromosomes of the organism.

2. Sexual reproduction.A process in which living organisms produce new individuals with genetic traits different from those of their parents.




2. Variant questions: 1. Sexual reproduction in the higher living organisms depends on two main processes; mention them? Then explain briefly. Why is this kind of reproduction considered a source of genetic variation?Sexual reproduction depends on two main processes which are gametes formation and fertilization.It is considered as a source of genetic variation because the offspring resulting from sexual reproduction combines the genetic traits from two sources.

2. Study the opposite figure, then answer the questions:a. What does this figure represent?Sperms surround the ovum before fertilization.

b. What is the number of chromosomes in each part in the figure?(N)

c. What is the name of the resulted structure and the number of chromosomes in it?Zygote – (2N).

d. Mention the characteristics of the resulted structure?The resulted structure has new genetic traits that combine its parent traits.

3. Give reasons for: 1. Asexual reproduction in plants does not need the presence of gametes.Because it is a vegetative reproduction occurs by the plants organs (leaves, roots and stems).

2. Asexual reproduction produces offspring with genetic traits identical to those of their parents.Because it occurs through one father individual and through a mitotic division as the new individual gets a genetic copy identical to the father.

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