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suter [353]
4 years ago
6

Consider the model above. It represents the electrical force. As r increases, the attractive force decreases. How would this mod

el compare to a model for the force of gravity?
Physics
2 answers:
kykrilka [37]4 years ago
7 0

Answer:

The answer is C

Explanation:

I did it on usatestprep and got it right

aivan3 [116]4 years ago
6 0

Answer:

As we keep on increasing the radius the value of the gravitation force of attraction decreases and as we decrease the radius the gravitation force increases.

Explanation:

Like the coulombs law of electrostatics, the law of gravitation also depends inversely on the square of the value of r. Therefore, as we keep on increasing the value of r the value of the gravitation force decreases and as we decrease the value of the r the value of gravitation force increases.

Gravitation Force=\frac{Gm_{1}m_{2} }{r^{2}}

Coulombs's Law= \frac{Kq_{1}q_{2} }{r^{2}}

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3 years ago
Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el
rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

q_1 = 21.0\mu C

q_1 = 47.0\mu C

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

E = \frac{kq}{x}^2

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2

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\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}

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E = 3.84*10^{-4} N/C

6 0
3 years ago
Read 2 more answers
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