Answer:



Explanation:
= Torque = 36.5 Nm
= Initial angular velocity = 0
= Final angular velocity = 10.3 rad/s
t = Time = 6.1 s
I = Moment of inertia
From the kinematic equations of linear motion we have

Torque is given by

The wheel's moment of inertia is 
t = 60.6 s
= 10.3 rad/s
= 0

Frictional torque is given by

The magnitude of the torque caused by friction is 
Speeding up

Slowing down

Total number of revolutions


The total number of revolutions the wheel goes through is
.
Answer:
<em> The distance required = 16.97 cm</em>
Explanation:
Hook's Law
From Hook's law, the potential energy stored in a stretched spring
E = 1/2ke² ......................... Equation 1
making e the subject of the equation,
e = √(2E/k)........................ Equation 2
Where E = potential Energy of the stretched spring, k = elastic constant of the spring, e = extension.
Given: k = 450 N/m, e = 12 cm = 0.12 m.
E = 1/2(450)(0.12)²
E = 225(0.12)²
E = 3.24 J.
When the potential energy is doubled,
I.e E = 2×3.24
E = 6.48 J.
Substituting into equation 2,
e = √(2×6.48/450)
e = √0.0288
e = 0.1697 m
<em>e = 16.97 cm</em>
<em>Thus the distance required = 16.97 cm</em>
Answer:
It’s called a conservative field.
Explanation:
I think it’s going to be the conservative field because in the question it talks about how it is able to become possible to define potential at a point in an electric field because electric field.
Answer:
In the words of Hartshorn and Alexander: “Economic Geography is the study of the spatial variation on the earth’s surface of activities related to producing, exchanging and consuming goods and services. Whenever possible the goal is to develop generalizations and theories to account for these spatial variations.”
Explanation: