Physics Semester 2 Review and Tutorial

SEMESTER 2 REVIEW

&TUTORIAL

Slides with red backgrounds involve word problems.

Slides with tan backgrounds involve matching concepts.

Slides with olive backgrounds involve reading data tables.

Slides with green backgrounds involve graphing.

The slides used in this tutorial are color coded. If you are experiencing difficulty

with one aspect of your understanding than another you might find this coding useful.

Make sure you know how to use the following formulas.These will be given to you on a separate card in finals day.

Displacement, Time, Velocity, Acceleration, Mass, Force, Weight, Tension, Impulse, Momentum, Work, Energy, Power, Spring Constant, Radius, Tangential Velocity, Centripetal Acceleration, Centripetal Force, Angular Momentum, Wavelength, Frequency, Time Period, Mach Number, Index of Refraction

MatchSelect the units that match the physics quantity.

cycles/s

kgm2/s2

J/s kg kgm/s

kgm2/s

m m/s m/s2 N or

kgm/s2

N/m None Ns s s/cycle

Displacement- mTime- sVelocity- m/sAcceleration- m/s2

Mass- kg Force- N Weight- N Tension- NImpulse- NsMomentum- kgm/sWork- JEnergy- J

Matched

Power- J/sSpring Constant- N/mRadius- mTangential Velocity- m/sCentripetal Acceleration- m/s2

Centripetal Force- NAngular Momentum- kgm2/sWavelength- mFrequency- Hz (cycles/s)Time Period- s/cycle or just s Mach Number- NoneIndex of Refraction- None

Analyze the Energy Bar Graph for a 1000 kg rollercoaster. Calculate the height of each hill.

Work Ug KE Ug KE Ug KE Ug KE0

100000

150000

200000

250000

Energy Bar Graph

Analyzed Graph

20.41 m

15.31 m

17.86 m

100000

150000

200000

250000

Energy Bar Graph

Ug = mgΔy

Analyze the Energy Bar Graph for a 1000 kg rollercoaster. Calculate the velocity on each hill.

100000

150000

200000

250000

Energy Bar Graph

Analyzed Graph

10.00 m/s

17.32 m/s

7.08 m/s

100000

150000

200000

250000

Energy Bar Graph

KE = ½ mv2

Solve for Force

What is the net forcerequired to get therollercoaster to the topof the first hill according to theprevious energy bar graph?

SOLUTION:

Find the force.

W = 200,000 J FΔy = 20.41 m

F = 9799.12 N

Solve for kinetic energy.

What is the kinetic enerycontained in a 35 kg object

thathas been displaced 7 metershorizontally by a force of 6 N?

SOLUTION:

Find the kinetic energy.

F = 6 N KEΔx = 7 m

KE = 42 J

Solve for units.

If that same 35 kg object wasdisplaced 7 meters horizontally

bya force of 6 N in 2 seconds,

whatwould be the units of your

answer?

W = J Unitst = s

Units = J/s

Find units.

P = W/t

EnergyImpulseMomentumPowerWork

• The product of force and the time interval during which the force acts.

• The product of the constant force on an object and the straight line distance through which the object is moved.

• Rate at which work is done.• Inertia in motion.• The ability to do work.

Energy- The ability to do work.Impulse- The product of force and the time interval during which the force acts.Momentum- Inertia in motion.Power- Rate at which work is done.Work- The product of the constant force on an object and the straight line distance through which the object is moved.

Matched

Analyze the graph. Determine the magnitude and units of the slope.

0 5 10 15 20 25 30 35 40 450

Force vs Change of Length

Δ l (m)

Analyzed graph.

0 5 10 15 20 25 30 35 40 450

Δ l (m)

Rise = 15 NRun = 25 mSlope = Rise/RunSlope = .6 N/m

Analyze the graph. Determine the units of the area.

0 5 10 15 20 25 30 35 40 450

Δ l (m)

Analyzed graph.

0 5 10 15 20 25 30 35 40 450

Δ l (m)

Area =( ½) Rise x RunArea = N x mArea = kgm/s2 x mArea = kgm2/s2

Area = J

Determine where the yintercept on a force vs

changeof length graph would be if

thespring contains no energy.

Determine y-intercept

SOLUTION:

Determine the y-intercept.

Us = 0 J y-intercept Δl = 0 m

y-intercept = 0,0

Two identical 924 kg cars begin breaking

at exactly the same time with the same

constant force of 1250 N on a level road.

Car “A” comes to a stop in 50 meters. Car

“B” comes to a stop in 100 meters.Determine the velocity of each

vehicle.

Solve for velocity.

SOLUTION:

To solve for velocity remember the Work-Kinetic Energy Theorem

F = 1250 N Vf”A”

Xi = 0 m Vf”B”

Xf”A” = 50 mXf”A” = 100 mm = 924 kg Vf”A”= 11.63 m/s Vf”B”= 16.45 m/s

Work Us KE Us KE Us KE0

10,000

12,000

Energy Bar Graph

Analyze the Energy Bar Graph. Determine what is happening to this energy transfer over time.

Analyzed graph. Energy is being transferred from a spring to a moving object.

Work Us KE Us KE Us KE0

10,000

12,000

Energy Bar Graph

Solve for velocity

A stopped car was leftunattended on a 17 meterhill. It rolls down hill for 5meters. Determine thevelocity of the car.

SOLUTION:

Determine the velocity.

Yi = 17 m V Yf = 12 m Δy = 5 m

V = 9.9 m/s

Analyze the graph. Choose the part of the graph best represents the picture.

Work KE Us Ug KE Us Ug KE Us Ug KE Us Ug KE Us Ug KE Us Ug0

Energy Bar Graph of a Pole Vaulter

Analyzed graph.

Analyze the graph. Choose the part of the graph best represents the picture.

Analyzed graph.

Solve for frequency

Determine the frequency that anobject is moving around a 30meter diameter circle if it istraveling at 5 m/s.

SOLUTION:

Calculate the frequency.

r = 15 m TV = 5 m/s f

f = .053 Hz

You could also use 2πr as λ!

Radius (m)

Time Period

Centripetal

Acceleration (m/s2 )

Centripetal

Force (N )

Frequency (Hz)

1 .898

4 12.25

13 .086

Complete the data table of a 2 kg object moving at a constant tangential velocity of 7 m/s.

Completed data table.Radius

(m)Time

Period(s)

Centripetal Acceleratio

n (m/s2 )

Centripetal

Force (N )

Frequency (Hz)

1 .898 49 98 1.11

4 3.59 12.25 24.5 .279

7 6.28 7 14 .159

13 11.63 3.77 7.54 .086

Frequency (Hz)

TimePeriod

Centripetal Acceleratio

n (m/s2 )

Centripetal Force (N )

.2 37.9

.111 35.1

Complete the data table of a 3 kg object moving around a constant radius of 24 meters.

Completed data table.Frequency

(Hz)Time

Period(s)

Centripetal Acceleration

(m/s2 )

Centripetal Force (N )

.5 2 236.87 710.61

.2 5 37.9 113.7

.111 9 11.7 35.1

.077 13 5.61 16.83

Select the graph that best represents the inverse square relationship between centripetal acceleration and radius if ac

(m/s2) is plotted on the y-axis and r (m) is plotted on the x-axis.

Graph Options

Matched Graph Options

Select the graph that best represents the inverse square relationship between centripetal acceleration and radius if ac

(m/s2) is plotted on the y-axis and r (m) is plotted on the x-axis.

Caramel, a 1.9 × 1030 kg star was just discovered inside the Milky Way. Determine the force of attraction between the star and its orbiting planets.

Completed data table. Fg= G (m1m2)

Longitudinal WaveTransverse WaveReflected WaveRefracted Wave

• Bending of an oblique ray of light when it changes velocity due to a change in the medium in which it is traveling.

• Wave in which the individual particles of a medium vibrate back and forth in the direction in which the wave travels.

• Return of light from a surface in such a way that the angle at which the ray is retruned is equal to the angle at which it strikes the surface.

• Wave in which the individual particles of a medium vibrate back and forth perpendicular to the direction in which the wave travels.

Matched

Longitudinal Wave- Wave in which the individual particles of a medium vibrate back and forth in the direction in which the wave travels.Transverse Wave- Wave in which the individual particles of a medium vibrate back and forth perpendicular to the direction in which the wave travels.Reflected Wave- Return of light from a surface in such a way that the angle at which the ray is retruned is equal to the angle at which it strikes the surface.Refracted Wave- Bending of an oblique ray of light when it changes velocity due to a change in the medium in which it is traveling.

Analyze each of these sound waves. Which of these is longitudinal? Which is quieter. Which is lower pitched? Which contains more amplitude (energy)? Which has the longest wavelength?

“A” “B”

Analyzed graph.

Longitudinal? “A” and “B” because they are both sound waves!Quieter? “A”Lower Pitched? “B”More Energy? “B”Longest Wavelength? “B”

Wavelength

Amplitude

Frequency

Solve for speed of sound

Determine the speed of sound at a

temperature of 23.25° C.

SOLUTION:

Calculate speed of sound.

T = 23.5° C Vsos

Vsos = 343.95 m/s

Solve for frequency when you’re standing still and the temperature is zero.

Determine the frequency of sound

you will hear when the ice creamtruck is coming toward you and

thenwhen it is traveling away from

you ifthe truck emits a bell at a

frequency of244 Hz and is traveling at 2 m/s.

SOLUTION:

Calculate frequency.

fi = 244 Hz ff

V = 2 m/s Towards ff = 245.48 Hz

Away ff = 242.53 Hz

Who buys ice cream in that kind of weather???

Solve for frequency

Compare the frequencies of a sound

that has a .75 meter wavelength when

traveling in a temperature of -10° C

and the same sound wave when the

temperature is 10° C.

SOLUTION:

Calculate frequencies.

T = -10° C ff

T = 10° C -10 C f = 432 Hz

10 C f = 448 Hz

Match Graph Options

Select the graph that best represents the position of a sound wave that echoes off the ocean floor and returns to the source at the surface of the ocean. The ship creating the sound is the reference point.

Graph Options

Select the graph that best represents the position of a sound wave that echoes off the ocean floor and returns to the source at the surface of the ocean.

Matched

Point at which sound wave hits ocean floor.

Solve for velocity.

It is a beautiful 28° C day without

a cloud in the sky and you arewatching the air show on

LakeMichigan. A plane is traveling

atMach 2.5. Determine the

plane’svelocity.

SOLUTION:

Calculate velocity.

T = 28° C VMach = 2.5

V = 867 m/s

Find the velocity of sound for that day and then multiply that number by 2.5!

Solve for time.

If the same plane is flying 300 meters

off the ground, determine the time it

takes after the plane passes directly

overhead for you to hear the sonic

boom, and how far the plane would

travel in that same amount of time.

SOLUTION:

Calculate time and displacement.

T = 28° C tMach = 2.5 Xf

Vp = 867 m/sVs = 346.8 m/s t = .865 s

Xf = 749.96 m

Solve for wavelength

Determine the wavelength of asound wave that has a frequencyof 128.75 Hz and is traveling at 330 m/s.

SOLUTION:

Calculate wavelength.

f = 300 Hz λV = 769 m/s

λ = 2.56 m

UmbraPenumbraAdditive Primary ColorSubtractive Primary ColorComplementary Color

• Three colors of light absorbing pigments that when mixed in certain proportions will reflect any color of the spectrum.

• A partial shadow that appears where some of the light is blocked and other light can fall.

• Any two colors of light that when added together produce white light.

• Darker part of a shadow where all light is blocked.

• Three colors of light that when added together in certain proportions will produce any color of the spectrum.

Matched

Umbra- Darker part of a shadow where all light is blocked.Penumbra- A partial shadow that appears where some of the light is blocked and other light can fall.Additive Primary Color- Three colors of light that when added together in certain proportions will produce any color of the spectrum.Subtractive Primary Color- Three colors of light absorbing pigments that when mixed in certain proportions will reflect any color of the spectrum.Complementary Color- Any two colors of light that when added together produce white light.

Solve for velocity

Compare the velocities of a radio

wave and a sound wave when the

temperature is 17° C.

SOLUTION:

Calculate and compare velocities.

T = 17° C Vs

c = 3 x 108 m/s Vr = 3 x 108 m/s

Vs = 340.2 m/s

Radio waves are part of the electromagnetic spectrum!

Solve for velocity

Compare the change in velocities

of a sound wave and a radio wave

after hitting glass if the index of

refraction is 1.43 when thetemperature is 17° C .

SOLUTION:

Calculate and compare velocities.

T = 17° C Vs

c = 3 x 108 m/s Vr = 2.09 x 108 m/s

Vs = 340.2 m/s

Reflective velocity will equalIncidental velocity for sound.

Solve for frequency

Determine the frequency of aradio wave that has awavelength of 300 meters.

SOLUTION:

Calculate frequency.

c = 3 x 108 m/s fλ = 300 m

f = 1,000,000 Hz or 1,000 kHz or 100 MHz Did you remember radio waves are part of the electromagnetic spectrum?

Match Polarizer Options

Select the set of polarizers that would allow the most light to pass through.

Polarizer Options

Matched

Match Polarizer Options

Polarizer Options

Matched

Solve for index of refraction

Mr. Floyd is trying to separate the color

pink from white light by shining white

light through a diamond prism. If the

angle of incidence is 35° and the angle of

refraction is 13.7°, determine the index

of refraction for a diamond.

SOLUTION:

Calculate the index of refraction.

nair = 1 ndiamond Θi = 35°Θr = 13.7°

n = 2.42

Solve for image size

A person is 1.5 meters tall andis standing 5 meters in front ofa pinhole camera. The camerascreen is .1 meters from thepinhole. Determine the size ofthe image.

SOLUTION:

Calculate the size of the image.

So = 1.5 m Si p = 5 mq = .1 m

Si = .03 m

Solve for focal point.

A picture of a 1.5 meter objectproduces an image of 1.5 cm when the object is 4 metersfrom the camera. Determine the focal point of the camera.

SOLUTION:

Calculate the focal point.

So = 1.5 m f Si = .015 m qp = 4 m

f = .0396 m

Select the statements that match each type of mirror. Concave Plane

Upright Inverted Real Virtual Magnified Reduced Same size Reversed True (Not Reversed)

Concave- Upright or Inverted, Real or Virtual, Magnified, Reduced, or the same size, Reversed or True.

Plane- Upright, Virtual, Same size, and Reversed.

Matched

Select the statements that match a concave mirror.

The object is outside the center point.The object is at the center point.The object is at the focal point.The object is inside the focal point.

Upright Inverted Real Virtual Magnified Reduced Same size Reversed True (Not Reversed)

The object is outside the center

point.

Inverted, Real, Reduced, and

Reversed.

The object is at the center point.

Inverted, Real, Same size, and

Reversed.

The object is at the focal point.

No image is produced.

The object is inside the focal

point.

Upright, Virtual, Magnified, and

Matched

Solve for image size, image distance and type of image.

A 7.5 cm object is placed 10 cm infront of a concave mirror that hasfocal point of 20 cm. Determinethe image size and distance. Then,determine if the image is real orvirtual.

SOLUTION:

Calculate the image size and distance and type.

So = 7.5 cm Si p = 10 cm qf = 20 cm Real or Virtual

Si = -15 cm (meaning the image has flipped)

q = -20 cm (meaning the image is in the mirror)

Virtual

Black White Blue Cyan Green Magenta Red Yellow

Select the color that each object would appear if only red light was incident upon the objects.

Matched

Select the color that each object would appear if only red light was incident upon the objects.

Select the color that each object would appear if only cyan light was incident upon the objects.

Matched

Select the color that each object would appear if only cyan light was incident upon the objects.

Select the most likely reason that each cloud would appear the color illustrated.

-The cloud is made up of small sized particles that reflect high frequency waves.-The cloud is made up of medium sized particles that reflect medium frequency waves.-The cloud is made up of large sized particles that reflect low frequency waves.

The cloud is made up of small sized particles that reflect high frequency waves.

Matched

The cloud is made up of large sized particles that reflect low frequency waves.

The cloud is made up of medium sized particles that reflect medium frequency waves.

Solve for velocity

While playing the “milk bottle” game at the

amusement park, a .448 kg ball is thrown

at a constant horizontal velocity of 10.4

m/s and collides with a stationary .577 kg

milk bottle. If the two objects then stick

together, determine the velocity at which

they would continue to travel.

SOLUTION:

Find the velocity.

m1 = .448 kg Vf

m2 = .577 kg pg = -9.8 m/s2 Vi1 = 10.4 m/s Vi2 = 0 m/s

p = 4.66 kgm/s2

Vf = 4.55 m/s

Solve for impulse

Dr. Fiala, who has a

mass of 100 kg is traveling at a

constant velocity of 1.5 m/s.Determine the impulse felt by

theunfortunate freshman sittingstationary in their bumper car.

SOLUTION:

Find the impulse.

m = 100 kg pVi = 1.5 m/s I

I = 150 Ns

Solve for force

If the unfortunate freshman sitting

in the bumper car experienced the

impact for .03 seconds, determine

the force that Dr. Fiala applied to

their bumper car.

Ha, Ha, Ha, Ha

SOLUTION:

Find force.

m = 100 kg FVi = 1.5 m/sp = 150 kgm/s2

t = .03 sI = 150 Ns F = 5,000 N

Solve for time.

If it takes 85,000 W of power to

raise Sky Trek Tower requiring2,000,000 J of energy,

determine thetime required to lift the ride to

thetop.

SOLUTION:

Find time.

P = 85,000 W tET = 2,000,000 J t = 23.53 s

Solve for height.

Fully loaded the Sky Trek Towerhas a mass of 2349.28 kg. Determine the maximum

height ofthe ride.

SOLUTION:

Find height.

P = 85,000 W ΔyET = 2,000,000 Jt = 23.53 sm = 2349.28 kgg = 9.8 m/s2

Δy = 86.87 m

Solve for velocity.

Sky Trek Tower is fully enclosed to prevent objects

from falling out. If you did drop

your accelerometer out of thewindow by accident, determine

itsvelocity just before reaching

theground.

SOLUTION:

Find velocity.

P = 85,000 W VET = 2,000,000 Jt = 23.53 sm = 2349.28 kgg = -9.8 m/s2

Δy = 86.87 m V = -41.26 m/s

Vertical Accelerometer Readings

Select the vertical accelerometer reading that best matches the acceleration it would be experiencing on the fall from the Sky Trek Tower.

Matched

Force Diagra

Select the force diagram that best matches the reading on the previous vertical accelerometer.

Force Diagra

Matched

Select the force diagram that best matches the reading on the previous vertical accelerometer.

Solve for height.

To confirm the results from the

height slide, you decide to

triangulate the height of Sky TrekTower. Using a baseline of 20

meters,and a sightline height of 1.5

meters, you findθ1 (22°) andθ2 (20°).

Determine the triangulated height.

SOLUTION:

Find height.

b = 20 m hSLH = 1.5 mΘ1 = 22°Θ2 = 20°

h = 74.93 m

Solve for angle.

At a certain point in your ride on

your roller coaster your horizontal

acceleration is 14.53 m/s2. Determine the angle at which

yourhorizontal accelerometer

would beindicating.

SOLUTION:

Find angle.

a = 14.53 m/s2 Θg = 9.8 m/s2

Θ = 56°

Solve for g’s.

At a certain point in your ride on your roller coaster your

horizontal accelerometer has adeflection of 76°. Determine

thenumber of g’s being produced

atthat point.

SOLUTION:

Find g’s.

g = 9.8 m/s2 gΘ = 76° g = 4 g’s

Solve for acceleration.

At a certain point in your ride on your roller coaster your

vertical accelerometer is halfway

between the second and third line.

Determine your acceleration at that

point.

SOLUTION:

Find acceleration.

g = 9.8 m/s2 ag’s = 1.5 a = 4.9 m/s2

Force Diagra

msSelect the force diagram that best matches the acceleration you calculated for the previous problem.

Force Diagra

Matched

Select the vertical accelerometer reading that best matches the most force of support.

Matched

Select the vertical accelerometer reading that best matches the most force of support.

Select the vertical accelerometer reading that best matches the roller coaster traveling at constant velocity to the top of the first hill.

Matched

Select the vertical accelerometer reading that best matches the roller coaster traveling at constant velocity to the top of the first hill.

Select the vertical accelerometer reading that best matches the roller coaster traveling down the first hill.

Matched

Select the vertical accelerometer reading that best matches the roller coaster traveling down the first hill.

Select the vertical accelerometer reading that best matches the roller coaster actually accelerating at 4.9 m/s2.

Matched

Select the vertical accelerometer reading that best matches the roller coaster actually accelerating at 4.9 m/s2.

It has been my pleasure to spend the last 10 months investigating physics with you. I hope you have enjoyed the journey as much as I have. Have a great summer and if I don’t see you around the halls of GBN again, have a great life!

With Kindest

Regards,

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