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scoundrel [369]
3 years ago
8

Two charges q1 and q2, that are distance d apart , repel each other with a force of 6.40 N. what would be the force between two

charges q1 and= 2q and q2=3q, that are distance d apart?
Physics
1 answer:
slava [35]3 years ago
6 0

Q: Two charges q1 and q2, that are distance d apart , repel each other with a force of 6.40 N. what would be the force between two charges q1'=2q1 and q2'=3q2 that that are distance d apart?

Answer:

The force = 38.4 N

Explanation:

From coulombs law,

F = kq₁q₂/r² ............................ Equation 1

Where F = Force of attraction or repulsion between the charges, q₁ and q₂ = first and second charge respectively, r = distance between the charges, k = constant of proportionality.

When, F = 6.4 N, r = d m.

6.4 = kq₁q₂/d²......................... Equation 1

When q₁' = 2q₁, q₂' = 3q₂, r = d cm

F = k(2q₁)(3q₂)/d²

F = 6kq₁q₂/d².......................... Equation 2

Dividing Equation 1 by equation 2

6.4/F = kq₁q₂/d²/(6kq₁q₂/d²)

6.4/F = 1/6

F = 6.4×6

F = 38.4 N.

Thus the force = 38.4 N

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Answer:

The magnitude of the acceleration of the tip of the minute hand of the clock 1.675\times10^{-6}\ m/s^2.

Explanation:

Given that,

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a=1.675\times10^{-6}\ m/s^2

Hence, The magnitude of the acceleration of the tip of the minute hand of the clock 1.675\times10^{-6}\ m/s^2.

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