MY NOTES A spring with a mass of 2 kg has a damping constant 14 kg/s. A force of 3.6 N is required to keep the spring stretched
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The electric field created by Q at the position of q is .
The given parameters:
- <em>Magnitude of charge, = q</em>
- <em>Spherical charge, = Q</em>
- <em>Force experienced by both charges, = F</em>
The electric field created by Q at the position of q is calculated as follows;
where;
- <em>E is the magnitude of electric field strength </em>
- <em>F is the force experienced by both charges</em>
- <em>Q is the charge</em>
Thus, the electric field created by Q at the position of q is .
Learn more about electric field here: brainly.com/question/14372078
Answer : The total work done in raising the ball is, 0.98 J
Explanation : Given,
Mass of the ball = 0.075 kg
Height raised of the ball = 1.33 m
As we know that the object is moving with the constant velocity, that means the work done against the gravity will be the net-work done.
So, the work done will be:
where,
w = work done
m = mass of ball
h = height of ball
g = acceleration due of gravity =
Now put all the given values in the above formula, we get:
Thus, the total work done in raising the ball is, 0.98 J
To solve this problem we must keep in mind the concepts related to angular kinematic equations. For which the angular velocity is defined as
Where
Final angular velocity
Initial angular velocity
Angular acceleration
t= time
In this case we do not have a final angular velocity, then
Re-arrange for
Therefore the mangitude of the angular aceleration is 5449.1rad/s²