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3 years ago

(a) What is the acceleration due to gravity on the surface of the Moon?

1 answer:

7 0

<h2>a) Acceleration due to gravity on the surface of the Moon is 1.64 m/s²</h2><h2>b) Acceleration due to gravity on the surface of the Mars is 3.75 m/s²</h2>

Explanation:

a) Acceleration due to gravity

$g=\frac{GM}{r^2}$

G = 6.67 × 10⁻¹¹ m² kg⁻¹ s⁻²

Mass of moon, M = 7.35 × 10²² kg

Radius of moon, r = 1.73 × 10⁶ m

Substituting

$g=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{(1.73\times 10^{6})^2}=1.64m/s^2$

Acceleration due to gravity on the surface of the Moon is 1.64 m/s²

b) Acceleration due to gravity

$g=\frac{GM}{r^2}$

G = 6.67 × 10⁻¹¹ m² kg⁻¹ s⁻²

Mass of Mars, M = 6.418 × 10²³ kg

Radius of Mars, r = 3.38 × 10⁶ m

Substituting

$g=\frac{6.67\times 10^{-11}\times 6.418\times 10^{23}}{(3.38\times 10^{6})^2}=3.75m/s^2$

Acceleration due to gravity on the surface of the Mars is 3.75 m/s²

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Now

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From steam table

$h_2 = 2513.33 KJ/Kg$

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$P = 69654kW$

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Is there a certain context to this question so that I can answer it better?

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Suppose that in a lightning flash the potential difference between a cloud and the ground is 1.3×109 V and the quantity of charg

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Answer:

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B) v = 8420.75 m/s

Explanation:

We are given;

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Plugging in the relevant values gives;

ΔU = 30 × 1.3 × 10^(9)

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This means that the initial potential energy will be equal to the final kinetic energy since all the potential energy will be converted to kinetic energy.

Thus;

P.E = K.E

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Plugging in the relevant values;

3.9 × 10^(10) = ½ × 1100 × v²

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