Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 81 mN when sepa
rated by a distance of 2.86 m .The spheres are then brought together until they are touching, enabling the spheres to attain the same final charge q. After the charges on the spheres have come to equilibrium, they spheres are separated so that they are again 2.65 m apart. Now the spheres repel each other with a force of magnitude 12.15 mN.
a. What is the final charge on the sphere on the right?
b. What is the initial charge q1 on the first sphere?
a. What is the final charge on the sphere on the right?
b. What is the initial charge q1 on the first sphere?
1 answer:
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Answer:
Explanation:
<u>Uniform Acceleration </u>
When an object changes its velocity at the same rate, the acceleration is constant.
The relation between the initial and final speeds is:
Where:
vf = Final speed
vo = Initial speed
a = Constant acceleration
t = Elapsed time
It's known a train moves from rest (vo=0) to a speed of vf=25 m/s in t=30 seconds. It's required to calculate the acceleration.
Solving for a:
Substituting:
Answer:
81.9756 m/s
16.8 m
4.8795 Hz
Explanation:
m = Mass of string = 0.12 kg
L = Length of string = 8.4 m
T = Tension on string = 96 N
Linear density is given by
Spee of the wave is given by
The speed of the waves on the string is 81.9756 m/s
Wavelength is given by
The longest possible wavelength is 16.8 m
Frequency is given by
The frequency of the wave is 4.8795 Hz