Complete Question
The complete question is shown on the first uploaded image
Answer:
a
The tension is 
b
The time taken is 
c
The position for maximum velocity is
S = 0
d
The maximum velocity is
Explanation:
The free body for this question is shown on the second uploaded image
From the question we are told that
The mass of the bob is 
The angle is 
The length of the string is 
The tension on the string is mathematically represented as

substituting values


The motion of the bob is mathematically represented as

=> 
Where
is the angular speed
and
is the phase change
At initial position S = 0
So 

Generally
can be mathematically represented as

Where T is the period of oscillation which i mathematically represented as

So



substituting values


Looking at the equation

We see that maximum velocity of the bob will be at S = 0
i. e 
The maximum velocity is mathematically represented as

Where A is the amplitude which is mathematically represented as

So

Recall 
substituting values
Use the right equation. To calculate the normal force of an object at an angle, you need to use the formula: N = m * g * cos (x) For this equation, N refers to the normal force, m refers to the object's mass, g refers to the acceleration of gravity, and x refers to the angle of incline.
Answer:
A fast feather
Explanation:
The faster any item is, the more momentum it has
Complete Question:
A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.
The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.
Answer:
Power = 54.07 W
Explanation:
Mass of the block = 10 kg
Angle made with the horizontal, θ = 60°
Distance covered, d = 5 m
Tension in the rope, T = 40 N
Coefficient of kinetic friction, 
Let the Normal reaction = N
The weight of the block acting downwards = mg
The vertical resolution of the 40 N force, 





Power, 

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