21 looks right too
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Browse All TopicsHi guys, i'm preparing for the AP physics exam.... please help me answer these questions. thank you.
9. A diver initially moving horizontally with speed v dives off the edge of a vertical cliff and lands in the water a distance d from the base of the cliff. How far from the base of the cliff would the diver have landed if the diver initially had been moving horizontally with speed 2v ?
(A) d
(B) root(2d)
(C) 2d
(D) 4d
(E) It cannot be determined unless the height of the cliff is known.
10. Two bodies of masses 5 and 7 kilograms are initially at rest on a horizontal frictionless surface. A light spring is compressed between the bodies, which are held together by a thin thread. After the spring is released by burning through the thread, the 5?kilogram body has a speed of 1/5 meter per second. The speed of the 7?kilogram body is
(A) 1/12 m/s
(B) 1/7 m/s
(C) 1 / root(35) m/s
(D) 1/5 m/s
(E) 7/25 m/s
5. A 2?kilogram block rests at the edge of a platform that is 10 meters above level ground. The block is launched horizontally from the edge of the platform with an initial speed of 3 meters per second. Air resistance is negligible. The time it will take for the block to reach the ground is most nearly
(A) 0.3 s
(B) 1.0 s
(C) 1.4 s
(D) 2.0 s
(E) 3.0 s
6 A horizontal force F is used to pull a 5?kilogran block across a floor at a constant speed of 3 meters per second. The frictional force between the block and the floor is 10 newtons. The work done by the force F in 1 minute is most nearly
(A) 0 J
(B) 30 J
(C) 600 J
(D) 1,350 J
(E) 1,800 J
21. A square loop of copper wire is initially placed perpendicular to the lines of a constant magnetic field of 5 x 10-3 tesla. The area enclosed by the loop is 0.2 square meter. The loop is then turned through an angle of 90° so that the plane of the loop is parallel to the field lines. The turn takes 0.1 second. The average emf induced in the loop during the turn is
(A) 1.0 x 10-4 V
(B) 2.5 x 10-3 V
(C) 0.01 V
(D) 100 V
(E) 400 V
22. James Joule did much to establish the value of the
(A) universal gravitational constant
(B) speed of light
(C) mechanical equivalent of heat
(D) charge of an electron
(E) specific heat capacity of helium
23. An ideal gas in a closed container initially has volume V, pressure P. and Kelvin temperature T. If the temperature is changed to 3T. which of the following pairs of pressure and volume values is possible?
(A) 3P and V
(B) P and V
(C) P and V/3
(D) P/3 and V
(E) 3P and 3V
my answers are B B C A C D B
i really have no idea how to do some of them...
Thanks for helping.
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hi compscigirl,
9) i agree with avisit that C) is correct, if we can assume that the vertical velocity is constant over the complete way. if there is an vertical accelelration it's a different story. as i can't find any hint about this in the question i would agree.
5) if i get it right (possible translation problems :) ) the mass has no vertical spedd at the moment it gets tossed from the platform.
so the formula
s = (1/2)*g*t^2
can be used to determine the time. s=way of the mass, g=acceleration of gravity (9.81m/s^2, here youn can calculate with 10m/s^2), t=time needed for way
22) i agree with BigRat (nice tip)
23) agreed again with BigRat. if you heat up something it will extend. but what will happen i a 'closed container'?
more soon...
leflon
@leflon,
> i agree with avisit that C) is correct, if we can assume that the vertical velocity is constant over the complete way. if there is an vertical accelelration it's a different story. as i can't find any hint about this in the question i would agree.
If the vehicle was ever to hit the water, there would be a vertical acceleration because the initial vertical velocity of the vehicle was zero. The impact (vehicle with water) distance from the base of the cliff is directly proportional to the horizontal velocity of the vehicle (as avizit stated). Double the speed that the vehicle leaves the cliff, and it will land twice as far away.
ok, i will try it again (one last time)
10) the energy stiored in the spring is divided onto both bodies in an equal amount, resulting in two equally distributed cinetic energies.
so we have (1/2)*m1*v1^2 = (1/2)*m2*v2^2
doing a little bit of conversion you get
m1*v1^2 5kg*(1/5 m/s)^2 1 m
v2^2 = --------------------- -> v2=root(------------------
m2 7kg root(35) s
please tell me i'm not wrong again :-)
cheers
leflon
@ozo
is this
p' = m*v' has to be constant, as the initial system (no movement) has a momentum of 0 it has to be the same as the two bodies start moving. as the movements (velocities) are in oposite direction there resulting momentums add to 0.
-> (now in scalar mode) m1*v1=m2*v2 -> v2=(m1*v1)/v2 -> B)?
i should really start thinking before writing. i will try to improve.
thanks a lot
leflon
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by: ozoPosted on 2004-04-21 at 22:41:15ID: 10885714
You have 10 and 5 correct.