Answer:
Explanation:
Given
average speed of train
Maximum acceleration=0.05g
Now centripetal acceleration is


r=7346.93 m
(b)Radius of curvature=900 m
therefore 



Answer:
The work done on the suitcase is, W = 1691 J
Explanation:
Given data,
The force on the suitcase is, F = 89 N
The distance Russell dragged the suitcase, S = 19 m
The work done on the suitcase by Russell is equal to the work done on the suitcase to overcome the friction
The work done on the suitcase by Russell is given by the formula
W = F · S
Substituting the given values,
W = 89 N x 19 m
W = 1691 J
Hence, the work done on the suitcase is, W = 1691 J
Hey there!
the answer is
C. a tennis racket striking a tennis ball
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Answer:
20 cm
Explanation:
Given that a ball is released from a vertical height of 20 cm. It rolls down a "perfectly frictionless" ramp and up a similar ramp. What vertical height on the second ramp will the ball reach before it starts to roll back down?
Since it is perfectly frictionless, the Kinetic energy in which the ball is rolling will be equal to the potential energy at the edge of the ramp.
Therefore, the ball will reach 20 cm before it starts to roll back down.
Complete Question:
Metal sphere A has a charge of − Q . −Q. An identical metal sphere B has a charge of + 2 Q . +2Q. The magnitude of the electric force on sphere B due to sphere A is F . F. The magnitude of the electric force on sphere A due to sphere B must be:
A. 2F
B. F/4
C. F/2
D. F
E. 4F
Answer:
D.
Explanation:
If both spheres can be treated as point charges, they must obey the Coulomb's law, that can be written as follows (in magnitude):

As it can be seen, this force is proportional to the product of the charges, so it must be the same for both charges.
As this force obeys also the Newton's 3rd Law, we conclude that the magnitude of the electric force on sphere A due to sphere B, must be equal to the the magnitude of the force on the sphere B due to the sphere A, i.e., just F.