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3 years ago

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A meteoroid is in a circular orbit 600 km above the surface of a distant planet. The planet has the same mass as Earth but has a

radius that is 90 % of Earth’s (where Earth’s radius is approximately 6370 km ).The acceleration of the meteoroid due to the gravitational force exerted by the planet is most nearly

1 answer:

AVprozaik [17]3 years ago

6 0

<h2>Answer:</h2><h2>The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/ $s^{2}$ </h2>

Explanation:

A meteoroid is in a circular orbit 600 km above the surface of a distant planet.

Mass of the planet = mass of earth = 5.972 x $10^{24}$ Kg

Radius of the earth = 90% of earth radius = 90% 6370 = 5733 km

The acceleration of the meteoroid due to the gravitational force exerted by the planet = ?

By formula, g = $\frac{GM}{r^{2} }$

where g is the acceleration due to the gravity

G is the universal gravitational constant = 6.67 x $10^{-11}$ $m^{3} kg^{-1} s^{-2}$

M is the mass of the planet

r is the radius of the planet

Substituting the values, we get

g = $\frac{(6.67 * 10^{-11}) (5.972 * 10^{24}) }{5733^{2} }$

g = 12.12 m/ $s^{2}$

The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/ $s^{2}$

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#79

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$v_i = 0$

now the takeoff velocity is given as

$v_f = 300 km/h$

$v_f = 83.33 m/s$

acceelration is given as

$a = 2 m/s^2$

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Problem 9: Suppose you wanted to charge an initially uncharged 85 pF capacitor through a 75 MΩ resistor to 90.0% of its final vo

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Answer:

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$C\times 0.9E=CE(1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})$

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$-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}=-ln(10)$

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$t=75\times 10^6\ \times\ 85\times 10^{-12}\times ln{10}$

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