LL5 U2 Review - Physics€¦ · 0< $'9,&(7$.(

UNIT 2 TEST REVIEW

MY ADVICE

TAKE YOUR TIME!

Read Questions Carefully

Read all possible answer choices before choosing

Work Pad for Calculations-

if it is not working type your answer as a comment

Use degree mode!!!

I AM REALLY NICE….BUT

THE PRACTICE TEST AND TEST NUMBER/ORDER DO NOT MATCH

MAKE SURE TO COMPLETE PRACTICE TEST & CHECK THE KEY

BEFORE YOU TAKE UNIT 2 TEST!

OFFICE HOURSTUESDAY 9AM – 10AM

ORCALL/TEXT ANY QUESTIONS YOU

HAVE BEFORE TEST!

• Given: vi = 27.0m/s teammate=82.0m away

• Make sure you are in degree mode for all calcs!

• draw pic

23.

• Find horizontal and vertical velocity components

•

•

• = 21.7m/s = 16.1m/s

• Along the curve the change in V for vertical motion can be found by:

•

• = -16.1 – (16.1) • 𝑣 negative because downward motion on second half of curve

• = -32.2 m/s

• Find the time the ball is in the air

• solve for

•

•.

.

• 3.28 sec

• Find the horizontal displacement

•

• 21.7 (3.28)

• =71.2 m

• Teammate is 80m away. How far must he walk to catch ball?

•

• 8.8 m

24.

37.8 North

29.5 North

? South – remember headwind is direct in opposite direction

• Velocity measured at ground is the resultant velocity

• falcon velocity + wind velocity = 29.5 m/s North

• = 37.8 + (x) = 29.5

• x = 29.5-37.8

• = -8.30 m/s

• = 8.30 m/s due south

25.

• How do we find the westward motion?

• Subtract the east from the west.

• 1.2 x 103 -1.5

• 1198.5

• Label what we know

• Find displacement using Triangle Method

1,198.5 m

2.5 x102 m

• d =

•

•

• =1224.2

• =1.2 x 103 m North of West

displacement m

Add to

Notes

26.

• Given info:

• F = 148N

• mass = 544kg

• acceleration = ?

• We use

• F= ma

• Solve for a

• a =

• = 148/544

• = .27205

• a = .272 m/s2

• Given info:

• 1 brick = 8.0N

• .31

• 4 bricks = 32N

•

• Solve for

• =

• Is 8N a mass?

• It is actually the

• =

• = .31 x 32

• = 9.92 N

• = 9.9 N to begin motion

𝑣 = 12.5𝑚/𝑠

• Given info:

•

• - 9.81m/s2

• (vf)2 = (vi)2 + 2aΔx• Solve for Δx

•

• =

• = -156.25/-19.62

• = 7.9638

• = 7.96m

236,000,000,000,000,000,000 m

1,000 m

1 km= 236,000,000,000,000,000 km

236,000,000,000,000,000 km

= 2.36 x 1017 km7.

Interpreting Graphs

What does this graph show us?

It shows that the object is accelerating as the velocity increases.

Instantaneous Velocity is the found by the slope of the tangent line

10.

DON’T FORGETVECTOR ADDITION

TRIANGLE METHOD

• Tail of one vector is placed at the head of the other.

• The resultant is the vector drawn from the tail of the first to the head of the last vector.

12.

13.

DON’T FORGET

d1

d2

d3

d4

d5

d6

Which displacement vectors shown have vertical components that are equal?

d4 and d6

vertical component = x value

19.

20.MOTION OF BALL APPEARS

HOW??

To Player – It looks like it drops directly downward because

moving at same velocity (Only true for small velocities)

To Crowd – appears to drop downward and backwards from

players position

BRING ON THE AIR RESISTANCE

21.

Wind/Air Resistance

Force of Gravity

F=mam = 80kg

Fg = 784.8 N

Fg = 780 N

SCIENTIFIC NOTATION REVIEW

Operation Property Example

addition and subtraction power of 10 must have same exponent – adjust whole number then add*if number is greater than 9 must adjust back into correct notation

3 x 104

+ 8 x 104

11 x 104

= 1.1 x 105

multiplication am x bn = a x b m+n (2x104) x (8x109) = 16 x 1013

=1.6 x 1014

division am ÷ bn = a x b m-n 5.6 x 109 / 4.5 x 105

=1.2 x 104

• (2.99 + 2.20 x 103) x (5.26 x 10-2)

• (2.99+2200)

• (2.20299 x 103) x (5.26 x 10-2)

• 11.5877 x 101

• 1.16 x 102

22.