P-201 Physics Exam 3 Chapters 10-14 – Flashcards
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What is the Angular Speed (in rev/min) of the Earth as it orbits the Sun?
answer
SOLVE:
1 rev/year X 1 year/365 days X 1 day/24 hours X 1 hour/60 min
this gives you revolutions/minute bc the earth rotates around the sun in one year. (1 rev/year)
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As you drive down the road at 15m/s, you press on the gas pedal and speed up with a uniform acceleration of 1.02 m/s2 for .65 seconds. If the tires on your car have a radius of 33 cm, What is the angular displacement during this period of acceleration?
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theta= thetanotXt + 1/2 a t2
divided by the radius
theta= (15m/s)(.65)+.5(1.02)(.65^2) .33 cm converted to m
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Find the angular speed of the minute hand and the hour hand of the famous clock in London, England known as Big Ben
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w=2pi/60 radians per minute which then equals 2pi/3600 radians per second
for the minute hand:
2piX(1/12) /3600
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Express the angular velocity of the second hand on a clock in the following units
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1rev/60 sec X60 sec/1 minX 60 min/1hour
for rev/hour
1 rev/60 secX360 degrees/1 revX60 sec/1 min
for degrees/minute
1 rev/60 secX2pi/1rev
for rad/sec
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a spot of paint on a bicycle tire in a circular path with a radius of .33 m. When the spot has traveled a linear distance of 1.60 m , through what angle has the tire rotated? Give your answer in radians.
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theta=s/r
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find the Angular speed of the earth as it spins about its axis, Give your answer in rad/s.
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2pi/24 hours=2pi/86400s
7.2x10^5 is the answer. It makes one revolution in 24 hours, so simply change hours to seconds and revolutions to radians
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One of the most studied objects in the night sky is the Crab nebula, the remains of a supernova explosion observed by the Chinese in 1054. In 1968 it was discovered that a pulsar-a rapidly rotating neutron star that emits a pulse of radio waves with each revolution-lies near the center of the Crab nebula. The period of this pulsar is 33 ms. What is the angluar speed (rad/s) of the nebula?
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w=2pi/T
w=2pi/.033
=190rad/s
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A pulley rotating in the counterclockwise direction is attached to a mass suspended from a string. The mass causes the pulley's angular velocity to decrease with a constant angular acceleration -2.20 rad/s2f the pulley's initial angular velocity is 5.10 rad/s , how long does it take before the angular velocity of the pulley is equal to −5.0rad/s?
answer
w=deltaaXt
t=delta w/a
t=(-5-5.1)/-2.20
=4.59
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On a certain game show, contestants spin a wheel when it is their turn. One contestant gives the wheel an initial angular speed of 3.40 rad/s. The angular acceleration of the wheel is -0.736 rad/s2. Through what angle has the wheel turned when its angular speed is 2.80 rad/s ?
HELP HELP HELP HELP HELP HELP HELP HELP HELP HELP
answer
w^2=wnot^2 +2a(delta theta)
delta theta=(w^2-wnot^2)/2a
=(2.80-3.40)/2X -.736
answer=
=2.53???????
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The angular speed of a propeller on a boat increases with constant acceleration from 13 rad/s to 25 rad/s in 3.0 revolutions. What is the acceleration of the propeller?
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w=wnot +alphaXt
t=2X2piX 3 revolutions/w+wnot
alpha=w-wnot/t
=(25-13)/.99
=12.09 rad/s^2
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The angular speed of a propeller on a boat increases with constant acceleration from 12 rad/s to 28 rad/s in 2.4 seconds.Through what angle did the propeller turn during this time?
answer
alpha=delta w/t
alpha=(28-12)/ 2.4
=6.67
theta=wnotXt +.5 at^2
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At 3:00 the hour hand and the minute hand of a clock point in directions that are 90.0 ∘ apart. What is the first time after 3:00 that the angle between the two hands has decreased by half to 45.0 ∘ ?
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????????
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The Earth's rate of rotation is constantly decreasing, causing the day to increase in duration. In the year 2006 the Earth took about 0.840 s longer to complete 365 revolutions than it did in the year 1906.What was the average angular acceleration of the Earth during this time? Give your answer in rad/s2.
answer
Begin by converting
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When the Earth, Moon, and Sun form a right triangle, with the Moon located at the right angle, as shown in the figure(Figure 1) , the Moon is in its third-quarter phase. (The Earth is viewed here from above its North Pole.)
answer
begin by finding the distance from sun to earth using the Square root of (distance from earth to sun)^2 and (distance moon to earth)^2
this will give you the distance from the sun to the moon, which you need to find to find the overall force.
to find the overall force, use G(m1m2)/r2 for both the sun and the moon, then combine them and solve c
the radius here is the distance + the radius of the moon, +radius sun
