Answer:
The frictional force is 
Explanation:
From the question we are told that
The coefficient of kinetic force is μk = 0.35
The normal force felt by the puck is 
Generally the frictional force that acts on the puck is mathematically represented as

=> 
=> 
Answer:
A
B

C

D

Explanation:
Considering the first question
From the question we are told that
The spring constant is 
The potential energy is 
Generally the potential energy stored in spring is mathematically represented as 
=>
=>
=>
Considering the second question
From the question we are told that
The mass of the dart is m = 0.050 kg
Generally from the law of energy conservation

=> 
=> 
Considering the third question
The height at which the dart was fired horizontally is 
Generally from the law of energy conservation

Here KE is kinetic energy of the dart which is mathematical represented as

=> 
=> 
=> 
Considering the fourth question
Generally the total time of flight of the dart is mathematically represented as

=> 
=> 
Generally the horizontal distance from the equilibrium position to the ground is mathematically represented as

=> 
=> 
Answer:
1470kgm²
Explanation:
The formula for expressing the moment of inertial is expressed as;
I = 1/3mr²
m is the mass of the body
r is the radius
Since there are three rotor blades, the moment of inertia will be;
I = 3(1/3mr²)
I = mr²
Given
m = 120kg
r = 3.50m
Required
Moment of inertia
Substitute the given values and get I
I = 120(3.50)²
I = 120(12.25)
I = 1470kgm²
Hence the moment of inertial of the three rotor blades about the axis of rotation is 1470kgm²
Answer:
It is an example of velocity
Explanation:
It is an example of velocity Don't ask how I know because I do know it I just don't know how to explain it.
C. A 1200kg car is going 15m/s