For a body falling freely from rest (disregarding air resistance), the distance the body falls varies directly as the square o
f the time. If an object is dropped from the top of a tower 490 ft high and hits the ground in 7 sec, how far did it fall in the first 5 sec?
2 answers:
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a) 32 kg m/s
Assuming the spring is initially at rest, the total momentum of the system before the collision is given only by the momentum of the bowling ball:
The ball bounces off at the same speed had before, but the new velocity has a negative sign (since the direction is opposite to the initial direction). So, the new momentum of the ball is:
The final momentum after the collision is the sum of the momenta of the ball and off the spring:
where is the momentum of the spring. For the conservation of momentum,
b) -32 kg m/s
The change in momentum of bowling ball is given by the difference between its final momentum and initial momentum:
c) 64 N
The change in momentum is equal to the product between the average force and the time of the interaction:
Since we know , we can find the magnitude of the force:
The negative sign simply means that the direction of the force is opposite to the initial direction of the ball.
d) The force calculated in the previous step (64 N) is larger than the force of 32 N.
Answer:
5.634 N rightwards
Explanation:
qo = - 3 x 10^-7 C
q1 = - 9 x 10^-6 C
q2 = 10 x 10^-6 C
r1 = 7 cm = 0.07 m
r2 = 20 cm = 0.2 m
The force on test charge due to q1 is F1 which is acting towards right
According to the Coulomb's law
F1 = (9 x 10^9 x 9 x 10^-6 x 3 x 10^-7) / (0.07 x 0.07)
F1 = 4.959 N rightwards
The force on test charge due to q2 is F1 which is acting towards right
According to the Coulomb's law
F2 = (9 x 10^9 x 10 x 10^-6 x 3 x 10^-7) / (0.2 x 0.2)
F2 = 0.675 N rightwards
Net force on the test charge
F = F1 + F2 = 4.959 + 0.675 = 5.634 N rightwards
