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:
Fnet = 0
Explanation:
- Since the block slides across the floor at constant speed, this means that it's not accelerated.
- According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
- This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:
- In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:
⇒ 169 N + Fn = Fg = 216 N (3)
- This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:
- Fn = 216 N - 169 N = 47 N (4)