The ratio of the forces is 9.59×10^-4.
The transverse speed in terms of tension in the string (T) of the wave is,
ω/k=√T/μ
T=ω^2k$2μ
Substituting the values we get,
T=[(50.0rad/s)^2/(0.800m−1)^2]×(12.0×10−3kg/m)
≈46.9N
The ratio of the transverse force and the tension is,
F/T=0.045N/46.9N
≈9.59×10^-4
Hence, the ratio of the forces is 9.59×10^-4.
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Answer:
The work done on the canister is 15.34 J.
Explanation:
Given;
mass of canister, m = 1.9 kg
magnitude of force acting on x-y plane, F = 3.9 N
initial velocity of canister in positive x direction, = 3.9 m/s
final velocity of the canister in positive y direction,
The change in the kinetic energy of the canister is equal to net work done on the canister by 3.9 N.
ΔK.E =
ΔK.E
The initial kinetic energy of the canister;
The final kinetic energy of the canister;
ΔK.E = 29.79 J - 14.45 J
ΔK.E = = 15.34 J
Therefore, the work done on the canister is 15.34 J.
Most of the momentum is transferred to the ball on top. Since the collision in this situation is elastic, momentum is conserved, meaning the momentum of both balls before hitting the floor is equal to the momentum of both balls right after the collision.
Answer:
When the force is applied along the direction of motion of the object.
Explanation:
When the force is applied in the direction of the velocity vector of the object, then only the magnitude of the velocity changes and the direction of motion remains unchanged. For example is you push an already moving cart in the direction its moving, then you will only changed its speed and not its direction.