Find the moment of force of 60 Newton about an axis of rotation at distance 20 cm from the force
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This question involves the concepts of the equations of motion, kinetic energy, and potential energy.
a. The kinetic energy of the rocket at launch is "3.6 J".
b. maximum gravitational potential energy of the rocket is "3.6 J".
<h3>a. KINETIC ENERGY AT LAUNCH</h3>The kinetic energy of the rocket at launch is given by the following formula:
where,
- K.E = initial kinetic energy = ?
- m = mass of rocket = 0.05 kg
= initial speed = 12 m/s
Therefore,
K.E = 3.6 J
<h3>b. MAXIMUM GRAVITATIONAL POTENTIAL ENERGY</h3>
First, we will use the third equation of motion to find the maximum height reached by rocket:
where,
- g = -9.81 m/s²
- h = maximum height = ?
- vf = final speed = 0 m/s
Therefore,
2(-9.81 m/s²)h = (0 m/s)² - (12 m/s)²
h = 7.34 m
Hence, the maximum gravitational potential energy will be:
P.E = mgh
P.E = (0.05 kg)(9.81 m/s²)(7.34 m)
P.E = 3.6 J
Learn more about the equations of motion here:
Answer:
The maximum no. of electrons-
Solution:
As per the question:
Maximum rate of transfer of charge, I = 1.0 C/s
Time, t = 1.0 h = 3600 s
Rate of transfer of charge is current, I
Also,
Q = ne
where
n = no. of electrons
Q = charge in coulomb
I = current
Thus
Q = It
Thus the charge flow in 1. 0 h:
Maximum number of electrons, n is given by:
where
e = charge on an electron =
Thus