Answer:
The motion of the ball relative to the ground is stationary
The motion of the ball relative to the wagon is backwards
Explanation:
To describe the motion of the ball relative to the ground, we note that
Assuming the ball is perfectly round and rotate freely, then we have
Force on the ball due to motion of the wagon = 0 N,
Then by the law of motion, an object will remain at rest when no force is applied to it
Therefore, apart from rotation of the ball, it will remain no displacement relative to the ground.
The motion of the ball relative to the wagon
Relative to the wagon, the ball appears to be moving in the opposite direction to the wagon, that is backwards.
Answer:

Explanation:
Let the child is moving with speed same as the speed of water flow
So here the position of child with respect to flow must be zero
And if the boat start at an angle with the vertical
so its relative speed with flow of water is given as


now the time to reach the child is given as

so now we have

So the time to catch the child is given as


So distance moved by it in 0.104 h
distance moved by the boat in upstream direction given as


In y direction the displacement of boat is

net displacement of the ball is given as



The distance covered by the body is 114.3 m
Explanation:
The work done by a force exerted on an object is given by

where
F is the magnitude of the force
d is the displacement of the object
is the angle between the direction of the force and of the displacement
For the object in this problem, we have
F = 350 N is the force applied
is the work done
if we assume that the force is applied parallel to the motion of the object
Solving for d, we find the distance covered by the object:

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Answer:
There is nothing faster then light
Explanation:
Answer:
a = 4.9(1 - sinθ - 0.4cosθ)
Explanation:
Really not possible without a complete setup.
I will ASSUME that this an Atwood machine with two masses (m) connected by an ideal rope passing over an ideal pulley. One mass hangs freely and the other is on a slope of angle θ to the horizontal with coefficient of friction μ. Gravity is g
F = ma
mg - mgsinθ - μmgcosθ = (m + m)a
mg(1 - sinθ - μcosθ) = 2ma
½g(1 - sinθ - μcosθ) = a
maximum acceleration is about 2.94 m/s² when θ = 0
acceleration will be zero when θ is greater than about 46.4°