Complete Question
The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.
(a) What is its angular acceleration in revolutions per minute-squared
(b) How many revolutions does the engine make during this 20 s interval?
rev
Answer:
a

b

Explanation:
From the question we are told that
The initial angular speed is 
The angular speed after
is 
The time for revolution considered is
Generally the angular acceleration is mathematically represented as

=>
=> 
Generally the number of revolution made is
is mathematically represented as

=> 
=> 
According to Newton's second law of motion, Force is the product of mass and acceleration of the object.
So, F = m * a
Here, m = 210 Kg
a = 2.4 * 10⁵ m/s²
Substitute their values,
F = 210 * 2.4 * 10⁵ N
F = 504 * 10⁵ N
F = 5.04 * 10⁷ N
In short, Your Answer would be Option B
Hope this helps!
Dropping a bouncy ball and stretching a rubber ban.
Answer:
Components: 0.0057, -0.0068. Magnitude: 0.0089 m/s
Explanation:
The displacement in the x-direction is:

While the displacement in the y-direction is:

The time taken is t = 304 s.
So the components of the average velocity are:


And the magnitude of the average velocity is
