Physics Warm Up: Agenda Copy these assignemts into your binder December 2 WarmUp: Agenda/ Review of...
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Physics Warm Up: Agenda Copy these assignemts into your binder December 2 WarmUp: Agenda/ Review of buoyant force InClass: Archimedes Worksheet Homework: Read and take notes p358-362 DUE NEXT CLASS December 3-4 WarmUp: Specific Heat of Copper Lab: Specific Heat Homework: Answer q 5p374 DUE NEXT CLASS December 5-6 WarmUp: Heat of Fusion Lab: Heat of Fusion Homework: When the air temperature is 22.2°C many people find it a comfortable temperature, yet the same people often find a swimming in 22.2°C water too cold to be comfortable. Use specific heat to explain the reason for the difference in sensation of temperature of the air and the water. Explain it to somebody and bring in a written YOU FINISH WRITING DOWN THE HOMEWORK, PLEASE GET OUT YOUR BUOYANCY ORK FROM LAST WEEK, QUESTIONS 1 AND 2 P324 SO THAT WE CAN GO OVER T IONS
Physics Warm Up: Agenda Copy these assignemts into your binder December 2 WarmUp: Agenda/ Review of buoyant force InClass: Archimedes Worksheet Homework:
Physics Warm Up: Agenda Copy these assignemts into your binder
December 2 WarmUp: Agenda/ Review of buoyant force InClass:
Archimedes Worksheet Homework: Read and take notes p358-362DUE NEXT
CLASS December 3-4 WarmUp: Specific Heat of Copper Lab: Specific
Heat Homework: Answer q 5p374 DUE NEXT CLASS December 5-6 WarmUp:
Heat of Fusion Lab: Heat of Fusion Homework: When the air
temperature is 22.2C many people find it a comfortable temperature,
yet the same people often find a swimming in 22.2C water too cold
to be comfortable. Use specific heat to explain the reason for the
difference in sensation of temperature of the air and the water.
Explain it to somebody and bring in a written copy of your
explanation. WHEN YOU FINISH WRITING DOWN THE HOMEWORK, PLEASE GET
OUT YOUR BUOYANCY HOMEWORK FROM LAST WEEK, QUESTIONS 1 AND 2 P324
SO THAT WE CAN GO OVER THE QUESTIONS
Slide 3
Buoyant force p324 There was some difficulty with problems 1
and 2 of the homework last week. We didnt go over it in class.
Question 1: A piece of metal weighs 50.0N in Air. 36.0N in water,
and 41.0N in oil. Find the density of a. The metal Two pages back
are the equations used in the sample problem F net = ( f V f o V o
) g where V net is the apparent weight, and the subscripts f and o
stand for The fluid and the object respectively Because the object
is submerged, the volumes are equal and we can simplify F net = ( f
o ) Vg F B f Vg F g(object) o Vg = Use the ratio (you can find this
on p322 if you forgot where it comes from) and solve for the
density of the object ( o ). FBFB F g(object) f = o Substitute
values from the question 14.0N 50.0N 1.00g/cm 3 = 3.57g/cm 3 =
3.57x10 3 kg/m 3
Slide 4
Buoyant force p324 There was some difficulty with problems 1
and 2 of the homework last week. We didnt go over it in class.
Question 1: A piece of metal weighs 50.0N in Air. 36.0N in water,
and 41.0N in oil. Find the density of Then B asks you to find the
density of the oil. F B f Vg F g(object) o Vg = We can use the same
simple ratio, but now solve for the density of the fluid, ( f ) =
640x10 3 kg/m 3 o F B F g(object) f = Substitute values from the
question 3.57x10 3 kg/m 3 9.0N f = 50.0N
Slide 5
Archimedes Principle Question 2 from that homework is easy if
you just remember that the buoyant force acting on a submerged
object is equal to the weight (force) of the fluid it displaces.
Verify it: On the demo table is a can with a spout. Fill the with
just enough water to run out of the spout. Weigh the small beaker.
When the spout stops dripping, put the beaker under it. Add the
100g weight to the water and collect the displaced water. Weigh the
beaker with the water and calculate the weight of the displaced
water. Record it on your paper. Should equal the difference in
weight The mass of this water Weigh the 100g weight in air by
hanging it from the hook on the balance. Fill the 250mL beaker and
place it on the arm of the balance. Weigh the 100g weight in water
and calculate the difference. The difference is the buoyant force.
It should be very close to the weight of the water displaced.
Slide 6
Buoyant force p324 2. An empty rubber balloon has a mass of
0.0120kg. The balloon is filled with helium at 0C, 1atm pressure
and a density of 0.181kg/m 3. The filled balloon has a radius of
0.500m. a. What is the buoyant force acting on the balloon The
buoyant force acting on the balloon is equal to the weight of the
air it displaces (p320). The mass of the displaced air is the
volume of the balloon times the density of the. Multiplying that by
acceleration due to gravity gives its weight. F g(air) = v f g F
g(air) =0.524m 3 1.29kg/m 3 9.8 m/s 2 F g(air) = 6.62kgm/s 2 =
6.62N V= 4/3 r 3 V= 4/3 (0.5m) V=.524m 3 B. The net force will be
the difference between the balloons weight and the buoyant force.
The weight of the balloon is the density of the helium times the
volume of the balloon added to the mass of the empty balloon. B. F
g(balloon) =9.8 m/s 2 (0.524m 3 )(0.181kg/m 3 )+9.8 m/s 2
(0.0120kg) F g(balloon) =1.05N F net = F g(air) - F g(balloon) F
net = 6.62N- 1.05N F net = 5.57N
Slide 7
Pressure & Pascals Law Pressure is force per area. p = f/a
Liquid Pressure = (depth)(density) p = h Units of pressure: pascal
= n/m 2 Atmosphere, mm of Hg (in manometer) kg/cm 2 (mass units).
Common pressures: 1 atm = 100 kPa 760 mm of Hg 10 m of H 2 O 1
kg/cm 2 Blaise Pascal Tap water pressure = 4 atm 400 kPa 4 kg/cm 2
Car tire 2 atm
Slide 8
Pressure contd Total Force = (pressure)(area) TF = pA.
Slide 9
Two Sample Problems Find the pressure needed to push water to
the top of Tower 2, a height of 8.0 meters. (Use mass units). The
density of water is 1.0 g/cm 3. Find the Total Force on the filled
school dam whose dimensions are 5.0m by 2.0m. The average depth is
1.0m. Mainmachinery Castle attack Solutions are on the next
page
Slide 10
Solutions Find the pressure needed to push water to the top of
Tower 2, a height of 8.0 meters. (Use mass units). The density of
water is 1.0 g/cm 3. Find the Total Force on the filled school dam
whose dimensions are 5.0m by 2.0m. The average depth is 1.0m.
12341234 p = h = (1.0m) (100cm/m)(1.0g/cm 3 2 ) = 100 g/cm 2 p = h
= (8.0m)(100cm/m)(1.0g/cm 3 2 ) = 800 g/cm 2 12341234 12341234 A =
LW = (5.0m)(2.0m) = 10.0 m 2 12341234 f = pA = ( 100g/cm 2 ) (10.0m
2 )(10 4 cm 2 /m 2 ) * = 1 X 10 7 g = 1 X 10 4 kg or 10 tons! Note:
1kg = 1000g 1 metric ton = 1000kg * There are 10 4 cm 2 in a m
2.
Slide 11
Pascals Law The pressure on a confined fluid is transmitted in
all directions. Pascals Vases show pressure depends only on depth
& density.
Slide 12
Pascal Pressure in Rocket (1) pascalrocket.mov
Slide 13
Hieros Fountain Demo h 2 is higher than h 1, so the pressure is
greater in the right system which pushes the water up into the
fountain. Try it! Add a about 50mL of water to the funnel of the
fountain on the demo table!
Slide 14
Hydraulic Lift: Demo: Syringes The pressure is transmitted
undiminished in all directions. Try this with different sized
syringes.
Slide 15
Hydraulic Brakes The applied pressure to the master cylinder is
transmitted equally to all four brake pistons.
Slide 16
Hydraulics lifts a House! (2) hydraulics.mov Oh, Pascal! Thanks
to Mark Shisler
Slide 17
An Uplifting Experience: Demo A strong rubber balloon inflated
beneath a car or truck can lift 15 metric tonnes of load. The
pressure is low, but the surface area is large. Total Force =
(press)(area). Demo: A garbage bag blown up by a vacuum cleaner can
lift a massive person.