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Artyom0805 [142]

2 years ago

12

If the distance between two charged particles is increased 2.69 times, the ratio of new to old electric force is _______ times t

he old value. (Round your answer to two decimal places).

1 answer:

Tasya [4]2 years ago

8 0

Answer:

F = K Q1 Q2 / R^2 force between 2 charged partices

F2 / F1 = (R1 / R2)^2 = (1 / 2.69)^2 = .139

F2 = .139 F1

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Satellite A is orbiting Earth at an altitude of 500 km and Satellite B is orbiting 800 km above the surface . How does the veloc

sp2606 [1]

The velocity of Satellite A is 2% greater than velocity of satellite B.

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<em>Altitude of Satellite B = 800 km</em>

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$F_c = \frac{mv^2}{r} \\\\F_g = \frac{GMm}{r^2} \\\\\frac{v^2}{r} = \frac{GM}{r^2}\\\\v^2 = \frac{GM}{r} \\\\v^2 r = GM\\\\v_1^2 r _1 = v_2^2 r_2\\\\v_A^2r_A = v_B^2 r_B\\\\(\frac{v_A}{v_B} )^2 = \frac{r_B}{r_A} \\\\\frac{v_A}{v_B} = \sqrt{\frac{r_B}{r_A} } \\\\\frac{v_A}{v_B} = \sqrt{\frac{(800,000\ + \ 6.4 \times 10^6}{(500,000\ + \ 6.4 \times 10^6} )} \\\\\frac{v_A}{v_B} = 1.02 \\\\v_A = 1.02 \ v_B$

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Learn more about velocity of satellite here: brainly.com/question/13981089

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3 years ago

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Sladkaya [172]

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3 0

2 years ago

(Please help 25 points)

Maksim231197 [3]

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<h3>The TE of Z is .... the TE of X</h3>

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2 years ago