4 years ago

NEED HELP!!!!!! CAN SOMEBODY HELP ME?

Physics

1 answer:

shepuryov [24]4 years ago

4 0

1) A
2) A
3) A
4) C
8) C
9) C
10) C
11) A
12) C
13) B
14) C
15) C
16) B
17) A

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Two charges, qA and qB, are separated by a distance, r, and exert a force, F, on each other. What new force will exist for the f

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Answer:

$\frac{1}{2}F$

Explanation:

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=9\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, at the beginning we have:

$q_1=q_A$ is the first charge

$q_2=q_B$ is the second charge

r is their initial separation

So the initial force is

$F=k\frac{q_A q_B}{r^2}$

Later, we have:

qA is doubled and r is doubled

This means that:

$q_A'=2q_A$ is the new charge

$r'=2r$ is the new separation

So the new force is:

$F'=k\frac{q_A' q_B}{r'^2}=k\frac{(2q_A)(q_B)}{(2r)^2}=\frac{1}{2}(k\frac{q_A q_B}{r^2})=\frac{1}{2}F$

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Answer:

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Explanation:

Since S=d/t, plug in the known values and solve

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0.25m/s*t=12m

t=12m/0.25m/s

t=48 seconds

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