Download Second Semester Final Exam Preparation 2012

Second Semester Final Exam Preparation Answers
Physics
Conceptual Review:
CENTRIPETALFORCE AND UNIVERSAL GRAVITATION
1. What is centripetal force? What effect does a centripetal force have on an object?
Centripetal force is an unbalanced force acting perpendicular to an object’s motion. The effect of the
centripetal force is to make the object accelerate by changing direction or turning. If the centripetal force
remains constant as the object turns in a complete circle, the center of the circle will always be in the
direction that the centripetal force is pointing.
2. You can make a simple device to detect acceleration (an accelerometer) by hanging a fishing weight on a
string under a protractor. While you speed up, the fishing weight will hang at an angle behind its support so
one components of the string’s tension will pull it forward to make it accelerate. While you slow down, the
fishing weight will hang at an angle in front of its support. What will happen when you drive in circles at a
constant speed? Why?
The weight angles toward outside of circle because it is accelerating toward the inside. Another way to say
it is that the string needs to angle so that one component of the tension pulls toward the center of the circle
and acts as the centripetal force to make the weight turn.
3. Everyone knows that there is no air on the Moon, but many people also imagine that there is no gravity on
the Moon. If you have ever watched video of the Apollo astronauts walking on the Moon, it is clear that the
Moon’s gravity holds the astronauts down. But even before we got to the Moon, we were positive there would
be gravity there.
a) What law of physics convinced us that the Moon had to have its own gravity?
Newton’s law of universal gravitation says that all massive objects create gravity and so the moon qualifies.
b) Could the Moon’s gravity affect us while we are still on the Earth?
Yes. According to Newton’s Law of Universal Gravitation, as you get further from an object, its gravity
gets weaker but will never completely go away (you can divide by a big enough distance to get zero). So
we are affected by the Moon’s gravity here on earth. As a matter of fact, it is the moon’s gravitation that is
responsible for the Earth’s tides.
MOMENTUM
4. During a game of disc golf, you are trying to throw your disc as far as possible when you throw your first
shot. To get the disc to go far, it needs to leave your hand traveling very fast. What can you do to make the
disc leave your hand going as fast as possible?
Another way to say that you want to the disc to go fast is that you want to give it a lot of momentum. To
give a lot of momentum, you need a big impulse. Since impulse is the product of force and time, a big
impulse comes from the biggest possible force multiplied by the biggest possible time.
5. Nerf balls are intended to be safer to use than another ball with the same mass. What is it about Nerf balls
that make them less likely to hurt you or a piece of furniture or a pet or anything else that they run into?
Whenever a ball hits anything, the ball’s target has to exert the exact right amount of impulse to stop the
ball. A Nerf ball is made from squishy foam, which means that it compresses to some degree whenever it
hits something. As it compresses, that gives more time for the impulse to occur. More time would mean a
greater impulse if the force stayed the same, but we need the exact right amount of impulse so the force
must decrease.
6. During a cartoon collision, Bugs Bunny stands still as he holds up his hand and stops a large artillery shell
from barreling into him. Why won’t this ever happen in the real world?
Even without exploding, an artillery shell is a fairly massive thing and it travels quite fast, which means that
it has a respectable amount of momentum. When the shell hits Bugs, momentum must be conserved which
means that during the collision the shell may lose momentum, but Bugs needs to gain the same amount of
momentum so the total remains the same. In other words Bugs should go flying backwards.
ENERGY
7. As a pickup truck glides at constant speed along flat ground, it does not do work on the sofa in the truck’s
bed. Explain how you can be sure that no work is done in this situation.
There are two ways to know that it does no work. The first is that the force the bed exerts on the sofa is
pointing upward (against gravity) while the sofa is moving forward. Since the force is not connected to the
motion, no work is done. The second is that the sofa’s energy does not change: it is moving at constant
speed, so it has the same amount of kinetic energy; it stays at the same height, so it has the same amount of
gravitational potential energy. Whenever work is done on object, it will gain or lose some form of energy
and that does not happen here.
8. Explain how energy is conserved when you load an egg into a slingshot, pull it back, point the slingshot
straight up and release the egg. Detail what happens to the egg and what happens to the energy during each
stage of the egg’s journey.
When you pull back the slingshot, you give it elastic potential energy as the cords are stretched.
When you let go, the cords shorten and the slingshot loses elastic potential energy, which is
transferred into kinetic energy in the egg as the egg moves faster. When the slingshot is back to its
original length, it has lost all of its elastic potential and the egg has gained it all as kinetic – so the egg is
moving at its highest speed.
As the egg flies upward, it will gradually gain gravitational potential energy as it gets higher and
higher above the ground. That gravitational potential has to come out of the only energy available, so
the egg will lose an equal amount of kinetic energy, which means it slows down continuously on the
way up.
It keeps losing kinetic energy until the kinetic energy is all gone and the egg has come to a stop. The
egg has the most gravitational potential energy that it can, so it is at its highest point above the ground.
With nothing to support it, the egg will immediately start to fall back to earth and will lose
gravitational potential energy as it gets closer to the ground. The lost gravitational potential energy
turns into kinetic, as the egg’s speed increases higher and higher.
This process ends when the egg collides with the ground at high speed. The egg comes to a stop as it
smashes into the ground and all of the kinetic energy is taken away. The collision converts the lost
kinetic energy into thermal energy, heating up the ground and the egg very slightly.
9. When we studied orbits, we treated the Earth’s orbit around the sun as a perfect circle. In reality the
Earth’s orbit is a slight ellipse, with the sun slightly off center (the picture is exaggerated).
a) Based on the law of conservation of energy, how would you expect the speed of earth in its orbit at aphelion
to compare to its speed at perihelion?
Since the earth is further away from the sun at aphelion, it will have greater gravitational potential energy.
In order for the earth’s total energy to be the same, it must have less kinetic energy and be going slower at
aphelion. (see pic on other page)
b) It turns out that the summer half of the year (from the vernal
equinox to the autumnal equinox) lasts almost a week longer
than the winter half of the year (from the autumnal equinox to
the vernal equinox) in the northern hemisphere [check a
calendar if you don’t believe it]. Why does our summer last
longer than our winter? [Hint: what part of our orbit are we in
during the summer? During the winter?]
Our summer lasts longer than our winter, because our
summer happens when we are further away from the sun.
That means that the earth is going slower in its orbit and has
a greater distance to travel – both of which mean that it will
take more time to make it around that portion of our orbit.
FLUIDS AND PRESSURE
9. How do drinking straws work? Why do we say there is no such thing as “sucking” in Physics?
When you are drinking with a straw, you begin by closing your mouth around the straw and making
the space inside your mouth as small as possible. When you increase the volume inside your mouth, the
pressure inside your mouth and the straw decrease. The pressure outside pushing down on the liquid
you are drinking is now much stronger then the pressure inside the straw, pushing the liquid up the straw
and into your mouth.
Pressure comes from the molecules of a fluid colliding with the objects around them. These collisions
always “push” the object, not “pull”.
10. Why do we use a knife to cut through butter and not the backside of a spoon?
The knife blade has a smaller area. From the equation P = F/A it is clear that using the same force on a
smaller area will create a larger pressure. It’s this increase in pressure that allows the knife to slice into
the butter.
11. Consider each situation. Decide whether the force of gravity, the buoyant force, and the pressure on each
block would be the same as or different than the other block for each of the following situations.
a) Two identical blocks of wood floating; one in water and one in mercury
Gravity is the same for both since the both have the same mass
Buoyancy force is the same for both since they are both floating and the force of gravity on both is equal
(remember: floating means Fg = FB).
Pressure is the same. This is easy to see, if we consider that the force of the fluid on the bottom surface
of the each block is equal to its weight and the area is the same for each block. Identical force divided
by the same area mean equal pressure on each block.
b) Two balls of equal size held underwater; one ball has a greater mass than the other ball
Gravity is greater for ball with larger mass
Buoyancy force is the same for both since they are displacing the same amount of water, and the density
of the fluid they are in is the same for both (FB = ρVg).
Pressure is the same since pressure is determined by the depth and the density of the fluid
(Pgauge = ρdg).
c) Two balls of equal size and mass; one held 10 m below the surface, the other held 5 m below
Gravity is the same for both since the both have the same mass
Buoyancy force is the same for both since they are displacing the same amount of water, and the density
of the fluid they are in is the same for both (Fbuoyancy = ρVg).
Pressure is greater for the ball that is deeper in the fluid, since pressure is directly related to depth.
(Pgauge = ρdg).
12. What is Archimedes’s Principle?
The buoyant force on an object is equal to the weight of the fluid that the object displaces. So the only
way for an object to get more buoyant force is to displace more of the fluid – this is familiar to anyone
who has loaded a boat, with each person getting into a boat it sinks deeper into the water, displacing
more water and getting more buoyant force.
13. How will the following affect the pressure of a fluid inside a container?
a) Increasing the volume of the container
PV = NkBT
Increasing the volume will decrease the pressure
b) Increasing the temperature inside the container
Increasing the temperature will increase the pressure
c) Increasing the number of molecules inside the container
Increasing the number of molecules will increase the pressure
14. Where does the buoyant force come from?
For any object in a fluid, the pressure below the object will be greater than the pressure above the object
since the bottom of the object is deeper in the fluid than the top of the object. This pressure difference
exerts an upward force on the object.
15. What is the pressure right at the surface of a body of water (where the depth is zero)? Where does this
pressure come from?
P = Pgauge + Patm
Pgauge = ρdg
Since d = 0m, the pressure from the water is zero.
However, we always have to add to this the pressure from the air, which is 101,300 Pa (this is a number
you need to have memorized!). You can think of us as living at the bottom of a huge ocean of air, and
all of those air molecules above us add up to a lot of pressure.
WAVES AND SOUND
16. What determines the period of a pendulum? What determines the period of a mass-and-spring system?
l
For a pendulum, (T = 2π
) the length of the pendulum is directly related to its period, and inversely
g
related to the acceleration due to gravity.
For a mass-and-spring system, (T = 2π
m
) the mass is directly related to its period, and inversely
k
related to the spring constant.
17. What is the difference between a longitudinal and a transverse wave? Give an example of both.
In a longitudinal wave, the vibration is parallel to the direction that the wave travels. In a transverse
wave, the vibration is perpendicular to the direction the wave travels.
Sound is a longitudinal wave, while light and ocean waves are examples of transverse waves.
18. What is it that “sets” the following wave properties?
a) Frequency
The source. The wave has the same frequency as the vibration that started the wave.
b) Velocity
The medium. All waves in the same
c) Wavelength
Both the source and the medium.
19. How do the frequency and amplitude of a sound wave affect how we hear that sound?
The frequency determines the pitch of the sound (higher frequency is higher pitch), and the amplitude
determines the volume (higher amplitude is louder).
20. Use the terms standing waves, natural frequencies, forced vibrations, and resonance to describe how
instruments work.
Most instruments consists of tubes or strings that have certain lengths. These lengths are directly related
to the wavelengths of the standing waves, as we can see in the equations L = nλ/2 and
L = nλ/4. These standing waves are those waves that will constructively interfere with their own
reflections, so they can have large amplitudes (and we hear them). Other wavelengths destructively
interfere with their own reflections, so they cancel themselves out and we don’t hear them. The
wavelengths of these standing waves correspond to the natural frequencies
(v = fλ). Forcing the instrument to vibrate at other frequencies will not create large amplitude
vibrations. Only the frequencies that you force the instrument to vibrate at that match up with the
instrument’s natural frequencies will create large amplitude vibrations and resonate.
21. a) What are the two ways a wave can reflect? What determines how a wave will reflect?
Inverted (if the end of the medium is fixed) and upright (if the end of the medium is free to vibrate).
b) What are the two ways that waves can interfere? How would the two types of interference between two
sound waves affect what we would hear?
Constructive (crest and crest or trough and trough) and destructive (crest and trough).
Constructive interference results in larger amplitude, meaning it would sound louder. Destructive
interference results in smaller amplitude, meaning it would sound quieter.
22. What is the difference between the standing waves in an open-open tube and an open-closed tube? How do
these standing waves affect the natural frequencies of the tubes?
For an open-open tube, L = nλ/2. For an open-closed tube, L = nλ/4.
If you have one of each type of tube, both the same length, the open-closed tube will have a lower
natural frequency (and a longer wavelength for the standing wave). The standing waves determine the
natural frequencies of the tubes.
23. How does the air-filled tube of a bassoon turn the many frequencies of the reed’s buzzing into a nice clean
tone? What happens to make the desired tone sound loudly? What happens to all of the other frequencies in
the reed’s buzzing?
The reed’s buzzing is made up of a range of frequencies. Each frequency of the buzzing sets up a
sound wave that travels into the bassoon’s tube and force the air inside there to vibrate at that frequency.
This means that the air inside the bassoon is vibrating with all of the same frequencies as the reed. But
there are some important differences between those vibrations.
Most of the frequencies create a simple forced vibration that makes the air inside the bassoon vibrate
with a small amplitude. Those small amplitude waves are quiet and hardly noticeable.
A small number of the reed’s frequencies match up with the natural frequencies of the air inside the
bassoon. That makes the air resonate with a large amplitude, which makes those sound waves loud and
easy to hear.
As you listen to a bassoon, you hear the loud resonance of the few sound waves that match the
bassoon’s natural frequencies and you do not hear the quiet forced vibration of all of the sound waves that
did not match the natural frequencies. The net result is a pleasant, harmonic sound from the bassoon.
Calculation Practice:
1 A 70 kg water skier holds onto the 30 m long tow rope as the boat drives in a tight circle and the skier swings
in a circle at the end of the rope at a constant speed of 20 m/s. How much tension does the rope exert on the
skier?
T = Fc = 933 N
2. NASA builds a gravitometer (a device for measuring the pull of gravity) to use in spaceships that will land
on other worlds. The gravitometer consists of a 0.00001 kg mass on the end of a fine spring that will stretch
when gravity pulls on the mass.
a) How much will the mass weigh if the spaceship lands on the surface of Mars?
Fg = 3.71 x 10-5 N
b) The spaceship lands on an unknown moon, whose radius is 1500 km and the mass on gravitometer weighs
6.5 x 10-5 N. How massive is the planet?
M = 2.19 x 1023 kg
3. A 0.5 kg basketball is being dribbled. It is heading downward at 15 m/s when it hits the floor and has a
force of 220 N push it upward until it bounces off the floor at 12 m/s. How long is the ball in contact with the
floor?
t = 0.061 s
4. One summer evening, sad because you haven’t learned any physics that day, you take a radar gun to the
carnival and watch the collisions of the bumper cars.
i) One collision you observe has a 160 kg bumper car heading east at 7.1 m/s run head on into a 180 kg
bumper car going 4.8 m/s to the west. After the collision, the first bumper car is heading west at 5.5 m/s.
a) How fast is the second car going?
vbf = 6.4 m/s to the east.
b) How much energy do the two cars have before the collision? [Hint: what kind of energy does each car
have? How much?]
Kai = 4032.8 J
Kbi = 2073.6 J
Ki total = 6106.4 J
c) How much energy do the two cars have after the collision? [Hint: what kind of energy does each car
have? How much?]
Kaf = 2420 J
Kbf = 3686.4 J
Kf total = 6106.4 J
d) Compare your answers to b) and c). Are they the same or different? Does this make sense? Why or
why not?
The total amount of energy is the same, so energy is conserved. Life is good.
ii) In another collision, a 150 kg car headed south at 7 m/s runs down the car in front of it, which was going 4
m/s also to the south. After the collision, both cars are still going south, the first car at 5 m/s and the second
car at 6.5 m/s.
a) How massive is the second car?
mb = 120 kg
b) How much energy do the two cars have before the collision?
Ki total = 5157 J
c) How much energy do the two cars have after the collision?
Kf total = 4410 J
d) Compare your answers to b) and c). Are they the same or different? Does this make sense? Why or
why not?
Since energy is conserved, the answers should be the same. The answers are not the same, which is a
problem. But it is a small problem, since other types of energy arise in collisions – notably thermal
energy. If the collision produces 747 J of thermal energy then all is right with the world.
5. You grab a piece of clothing on a clothesline and start walking away with it before it is completely clear of
the line. You pull with 30 N at 40 below horizontal for 0.6 m before realizing something is wrong.
a) How much work do you do on the clothesline?
W = 13.79 J
b) How much energy does the clothesline have when you are done pulling? What kind of energy?
The clothesline will bend and stretch so it has elastic potential energy.
Ue = 13.79 J
6. When you inherit some land, you decide to
build a frictionless roller coaster on it. The
300 kg rollercoaster car will start by
dropping off a high hill and then go through a
25 m diameter circular loop, after which it
will go over a 20 m high hill (as shown in the
picture)
a) How fast does the rollercoaster car
have to be going in order to stay on the
track at point B? (in other words, what is
the slowest possible speed for which it will make it around the loop without coming off?)
vB = 11.07 m/s
b) How high do you need to build the hill at point A in order for the rollercoaster car to have enough speed
to stay on the track when it gets to point B?
hA = 31.25 m
c) How fast will the rollercoaster car be going when it gets point C if it barely makes it round the loop at
point B?
vC = 14.85 m/s
d) How much normal force does the track exert on the car at point C?
FN = -367.5 N This is bad because normal force can’t pull downward. Looks like the car is coming off
the tracks
7. A 0.18 kg Hot Wheels car is placed into a spring loaded launcher and is pushed backwards10 cm with 30 N
of force. When the launcher is released, the car rushes out of the launcher and glides across the floor for 10 m
before coming to a stop. How strong is the force of friction against the car?
Ff = 0.3 N
8. a) Calculate the force of the atmosphere acting on the top of a table that measures 1.6 m x 2.9 m.
Patm = 101,300 Pa
A = 1.6 m x 2.9 m = 4.64 m2
P = F/A
or
F = P∙A
F = 101,300 Pa ∙ 4.64 m2
F = 470,032 N
b) What is the force of the atmosphere on the bottom of the table?
Since the air pressure and the area are the same, the force on the bottom of the table will also be equal to
470,032 N.
9. a) How much pressure does any one of the four legs holding up a 60kg loudspeaker exert on the floor, if the
legs are circular with a 2 cm diameter on the bottom?
For a circle, A = π∙r2 = π∙(0.01 m)2 = 3.141 x 10-4 m2
For 4 legs, that is 4 x 3.141 x 10-4 m2 = 1.257 x 10-3 m2
The legs have to exert the same amount force that gravity is pulling down on the loudspeaker
F = mg = 60 kg ∙ 9.8 N/kg = 588 N
P = F/A
P = (588 N) / (1.257 x 10-3 m2)
P = 467,915 Pa
b) How much pressure does a 1500 kg elephant standing on one foot (0.08 m2) exert on the floor?
P = F/A
P = (1500 kg ∙ 9.8 m/s/s) / 0.08 m2
P = 183,750 Pa
10. a) An oak wood block with a volume of 1.3 x 10-4 m3 and a mass of 0.10 kg is halfway submerged
underwater (density of water = 1,000 kg/m3).
i. What is the force of buoyancy on the block?
The block only displaces half of its volume: Vdisplaced = 1.3 x 10-4 m3 / 2 = 6.5 x 10-5 m3
FB = ρVg
FB = 1,000 kg/m3 ∙ (6.5 x 10-5 m3) ∙ 9.8 m/s/s
FB = 0.637 N
ii. Will the oak block float, sink, or rise at this point?
Fg = 0.10 kg ∙ 9.8 m/s/s
Fg = 0.98 N
Fg > FB
block will sink lower into the water
iii. What is the pressure on the block when it is at the surface?
P = Pgauge+ Patm
Pgauge = ρdg
Since d = 0m, Pgauge = 0 Pa.
P = Patm
P = 101,300 Pa
b) The oak block is then held fully submerged 3 m below the surface of the water.
i. What is the force of buoyancy on the block?
When submerged, the block is displacing a volume of water equal to its volume.
FB = ρVg
FB = 1,000 kg/m3 ∙ 1.3 x 10-4 m3 ∙ 9.8 m/s/s
Fbuoyancy = 1.274 N
ii. Will the oak block float, sink, or rise at this point?
Fg = 0.10 kg ∙ 9.8 m/s/s
Fg = 0.98 N
Fg < FB
block will rise
iii. What is the pressure on the block when it is 3 m below the surface?
P = Pgauge + Patm
Pgauge = ρdg
Pgauge = 1,000 kg/m3 ∙ 3 m ∙ 9.8 m/s/s = 29400 Pa
P = 29,400 Pa + 101,300 Pa
P = 130,700 Pa
11. a) Calculate the number of particles inside a helium-filled balloon that has a volume of 0.001 m3 if the
temperature inside the balloon is 22° C and the pressure inside is 130,000 Pa.
PV = NkBT or
N = PV / kBT
N = (130,000 Pa ∙ 0.001 m3) / (1.38 x 10-23 J/°K ∙ 295 °K)
N = 3.19 x 1022
b) If the temperature is raised 5° C and the volume remains constant, what is the pressure inside the balloon?
PV = NkBT or
P = NkBT / V
P = (3.19 x 1022 ∙ 1.38 x 10-23 J/°K ∙ 300 °K) / (0.001 m3)
P = 132,066 Pa
12. A tsunami wave with a frequency of 18 Hz travels at 120 m/s at it passes a warning buoy that is anchored
500 m from the shoreline.
a) What is the wavelength of the wave?
v = fλ or λ = v/f
λ = 120 m/s / 18 Hz
λ = 6.67 m
b) How much time do people on the beach have to get out of the way of the tsunami?
v = d/t or t = d/v
t = 500 m / 120 m/s
t = 4.17 s
13. Kelcie plays a flute, which works as an open-open tube. With all of the holes closed, it is 0.5 meters long.
She blows across the embouchure to create a hissing that includes all frequencies between 200 Hz and 1100 Hz.
a) What frequency or frequencies will you hear loudly as Kelcie plays?
The hissing sound is making the air inside the flute vibrate at all frequencies between 200 Hz and 1100
Hz, but most of the frequencies inspire really small amplitude vibrations in the tube. The exception is
the frequencies that match up with natural frequencies of the air inside the tube – they resonate and
vibrate with a big amplitude.
If we can identify the natural frequencies, we will know which ones vibrate with a big amplitude.
Fortunately, the natural frequencies are the ones that create standing waves, so all we need to do is find
the standing waves whose frequencies fall between 200 Hz and 1100 Hz.
Find the natural frequencies through the harmonic standing waves, one by one until we get to
frequencies higher than the original buzzing:
L = nλ/2 or λ = 2L/n
n =1
n=2
n=3
n=4
λ = 2∙(0.5 m)/1 = 1.0 m
λ = 2∙(0.5 m)/2 = 0.5 m
λ = 2∙(0.5 m)/3 = 0.33 m
λ = 2∙(0.5 m)/4 = 0.25 m
f= v/λ = (343 m/s)/1.0 m = 343 Hz
f= v/λ = (343 m/s)/0.5 m = 686 Hz
f= v/λ = (343 m/s)/0.33 m = 1029 Hz
f= v/λ = (343 m/s)/0.25 m = 1372 Hz
The fundamental, first overtone and second overtone all fall within the range that the hissing is
producing, so the 343 Hz, 686 Hz and 1029 Hz waves will resonate and sound loud. All other
frequencies are small amplitude and hard to hear.
b) If she were to close off the far end of the flute with her hand while continuing to blow the same way, what
frequency or frequencies of sound would you hear loudly?
The tube is now closed at one end while it remains open at the other, so it is a fixed-free oscillator.
Remember that fixed-free only have odd harmonics
L = nλ/4 or λ = 4L/n
n =1
n =3
n =5
n =7
λ = 4∙(0.5 m)/1 = 2.0 m
λ = 4∙(0.5 m)/3 = 0.667 m
λ = 4∙(0.5 m)/5 = 0.4 m
λ = 4∙(0.5 m)/7 = 0.286 m
f= v/λ = (343 m/s)/2.0 m = 171.5 Hz
f= v/λ = (343 m/s)/0.667 m = 514.5 Hz
f= v/λ = (343 m/s)/0.4 m = 857.5 Hz
f= v/λ = (343 m/s)/0.286 m = 1200.5 Hz
The second overtone and fourth overtone are all that fall within the range that the hissing is producing,
so the 514.5 Hz and 857.5 Hz waves will resonate and sound loud. All other frequencies are small
amplitude and hard to hear.
c) How does the pitch of the covered flute sound in comparison to the uncovered flute?
The covered flute is capable of making lower sounds than the uncovered flute, but because the first
harmonic is outside the frequencies that the player is exciting, it will not sound. This means that the
covered flute will sound higher because that 3rd harmonic of the covered flute (514.5 Hz) is higher than
the 1st harmonic of the uncovered flute (343 Hz).
14. A 714 Hz sound wave resonates with the 5th harmonic in an open-closed tube of air. How long is the tube?
v = fλ or λ = v/f
λ = (343 m/s/) / (714 Hz)
λ = 0.480 m
L = nλ/4
L = 5 ∙ (0.480 m) / 4
λ = 0.60 m
15. a) How long does it take sound to travel 1 mile (about 1600 m)?
v = d/t or t = d/v
t = 1600 m / 343 m/s
t = 4.66 s
b) How long does it take light to travel that same distance?
v = d/t or t = d/v
t = 1600 m / 3 x 108 m/s
t = 5.3 x 10-6 s
c) What is the time-delay between the moment you see lightning strike one mile away, and the moment you hear
the crack of thunder from that lightning strike?
4.66 s – 5.3 x 10-6 s = 4.66 s