A speeder goes by a police car at a constant speed of 20 m/s. Five seconds after
ID: 1515707 • Letter: A
Question
A speeder goes by a police car at a constant speed of 20 m/s. Five seconds after the speeder has passed the police car, the police car starts from rest with constant acceleration in the direction of the speeder. If the police car catches the speeder after accelerating for 20 s, what was the magnitude of the acceleration of the police car, in m/s^2? 9.8 8.2 6.4 4.1 2.5 A simple pendulum with a length of 1.0 m swings back and forth from a suspension point on the surface of a hypothetical Planet X. If the period of small-amplitude oscillations is 4.0 s and the radius of Planet X is 3.5 times 10^6 m. what is the mass of Planet X, in kg? 7.5 times 10^22 1.9 times 10^23 4.5 times 10^23 8.4 times 10^23 1.2 times 10^24 Two masses are attached to the ends of a massless string and the string is stretched over a massless, frictionless pulley, as shown. There is no friction between mass 2m and the incline. What is the acceleration of mass m (on the left), in m/s^2? 0.48, up 1.2, down 1.6, up 2.1, down 0Explanation / Answer
5 second after the speeder has passed the police car , the speeder will travel distance of
S1 = Vt = (20m/s)(5s) = 100m .
In rest of 20 s , the speeder will travel the distance of
S2= Vt = (20m/s)(20s) = 400 m
Total distance covered = S = S1+ S2 = 100m + 400m = 500m
Now using second equation of motion
S = Vit + 0.5 at2
Vi = 0m/s for police car
500 = 0.5 a t2
a = 500 / o.5(20)2
a = 2.5m/s2
option E , is the correct option.
B) The time period of the pendulum is given as
T = 2pi ( l / g)1/2
g = 4pi2 L / T2
g = 4(3.14)2 (1) / 42
g = 2.46m/s2
now by Newtons law of universal gravitation '
F = G M m / R2
as F = W = mg
mg = G M m / R2
g = G M / R2
2.46 = (6.67 * 10-11) M / (3.5 * 106)2
M = 4.5 * 1023 kg
option C is correct.
Please post next question in new post...you are asking to many questions in a single post.
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