The appropriate expression for the calculation of power by relating the angular energy in a given time.
In other words the instantaneous power of an angular accelerating body is the torque times the angular velocity

Where
Torque
Angular speed
Our values are given by


The angular velocity must be transformed into radians per second then


Replacing,



The average power delivered by the engine at this rotation rate is 211.1kW
I’m not too sure but I think it’s 8,91 m/s2
Yes,and because not everyone can wink and often that someone can only wink with one eye only
Explanation:
The weight of the car is equal to,
...........(1)
Where
m is the mass of car
g is the acceleration due to gravity
The normal or vertical component of the force is, 
or
.............(2)
The horizontal component of the force is, 
Taking ratio of equation (1) and (2) as :



or

Hence, this is the required solution.
Answer:
Explanation:
velocity of light in a medium of refractive index V = V₀ / μ
V₀ is velocity of light in air and μ is refractive index of light.
time to travel in tube with air = length of tube / velocity of light
8.72 ns = L / V₀ L is length of tube .
time to travel in tube with jelly = length of tube / velocity of light
8.72+ 1.82 = L / V L is length of tube .
10.54 ns = L / V
dividing the equations
10.54 / 8.72 = V₀ / V
10.54 / 8.72 = μ
1.21 = μ
refractive index of jelly = 1.21 .