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

=> 
=> 
The two spheres have opposite charges.
<h3 /><h3 /><h3>What are types charge?</h3>
- A charge can be negatively charged or positively charged.
- When two charges have opposite signs, that is positive and negative signs, the two charges will attract each other.
- When the two charges have the same sign, it causes repulsion.
When a positive charge points downwards ↓ and the negative charge points upwards ↑, this causes attraction and shows that the two charges are different.
Thus, we can conclude that the two spheres have opposite charges.
Learn more about attraction and repulsion of charges here: brainly.com/question/2396080
Answer:
I = 1.06886 N s
Explanation:
The expression for momentum is
I = F t = Δp
therefore the momentum is a vector quantity, for which we define a reference system parallel to the floor
Let's find the components of the initial velocity
sin 28.2 = v_y / v
cos 28.2= vₓ / v
v_y = v sin 282
vₓ = v cos 28.2
v_y = 42.8 sin 28.2 = 20.225 m / s
vₓ = 42.8 cos 28.2 = 37.72 m / s
since the ball is heading to the ground, the vertical velocity is negative and the horizontal velocity is positive, it can also be calculated by making
θ = -28.2
v_y = -20.55 m / s
v_x = 37.72 m / s
X axis
Iₓ = Δpₓ = 
since the ball moves in the x-axis without changing the velocity, the change in moment must be zero
Δpₓ = m
- m v₀ₓ = 0
v_{fx} = v₀ₓ
therefore
Iₓ = 0
Y axis
I_y = Δp_y = p_{fy} -p_{oy}
when the ball reaches the floor its vertical speed is downwards and when it leaves the floor its speed has the same modulus but the direction is upwards
v_{fy} = - v_{oy}
Δp_y = 2 m v_{oy}
Δp_y = 2 0.0260 (20.55)
= 1.0686 N s
the total impulse is
I = Iₓ i ^ + I_y j ^
I = 1.06886 j^ N s