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a_sh-v [17]
3 years ago
5

The acceleration due to gravity on the surface of Jupiter is about 2.5 times the acceleration due to gravity on Earth’s surface.

What would be the weight of a space probe on the surface of Jupiter?
A.2.5 times lighter than on Earth
B.6.25 times heavier than on Earth
C.2.5 times heavier than on Earth
D.6.25 times lighter than on Earth
Physics
1 answer:
Ann [662]3 years ago
6 0

Answer: The correct answer is option C.

Explanation:

Weight = Mass × Acceleration

Let the mass of the space probe be m

Acceleration due to gravity on the earth = g

Weight of the space probe on earth = W

W=m\times g

Acceleration due to gravity on the Jupiter = g' = 2.5g

Weight of the space probe on earth = W'

W'=mg'=m\times 2.5g

\frac{W'}{W}=\frac{m\times 2.5g}{m\times g}

W'=2.5\times W

The weight of the space probe on the Jupiter will be 2.5 times the weight of the space probe on earth.

Hence, the correct answer is option C.

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∆<em>W</em> = <em>M</em> <em>g h</em>

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Solve for <em>M</em>, then for <em>m</em> :

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And m, c, h are all constants from the cooling law relation

From Newton's law of cooling

Rate of Heat loss by the cake = Rate of Heat gain by the environment

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(d/dt) (T - T∞) = dT/dt (Because T∞ is a constant)

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Let (h/mc) be k

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Integrating the left hand side from T₀ to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -kt

(T - T∞)/(T₀ - T∞) = e⁻ᵏᵗ

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

Inserting the known variables

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(T - 70) = 142 e⁻ᵏᵗ

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kt = 0.4742 × 4 = 1.897

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T - 70 = 142 × 0.15 = 21.3

T = 91.3°F

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