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2 years ago

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One way to measure g on another planet or moon by remote sensing is to measure how long it takes an object to fall a given dista

nce. A lander vehicle on a distant planet records the fact that it takes 3.58 s for a ball to fall freely 12.02 m, starting from rest.
a. What is the acceleration due to gravity on that planet? Express your answer in m/s2.b. What is the acceleration due to gravity on that planet? Express your answer in earth g

1 answer:

Jlenok [28]2 years ago

5 0

Answer:

(a) 0.94 m/s²

(b) g (planet) = 0.096g

Explanation:

(a)

From Newton's equation of motion,

S = ut + 1/2gt²......................... equation 1

Making g the subject of equation 1

g =( S - ut)/t² ........................ equation 2

Where s = distance ( m), u = initial velocity (m/s), t = time (s), g = acceleration due to gravity (m/s²)

From the question, S = 12.02 m, t = 3.58 s, u= 0 ( at rest),

Substituting these values in equation 2

g = {12.02 -(0×3.58)}/3.58²

g = (12.02)/12.82

g = 0.94 m/s²

∴ The acceleration due to gravity on the planet = 0.94 m/s²

(b) g (planet)/g (earth) = 0.94/9.80

g (planet) = 0.096 g (earth).

The acceleration due to gravity of the planet in terms of the earth g is

g (planet) = 0.096g

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b) $\bigtriangledown T=1^{\circ}C.mile^{-1}$

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Given is the data of variation of temperature with respect to the distance traveled:

Temperature T as a function of distance d:

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(a)

Total change in temperature from the start till the end of the journey:

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$T_f$ = final temperature

$T_i$ = initial temperature

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So, correspondingly we have the eq. (2) & (1) as:

$\Delta T= (100+30)-(0+30)$

$\Delta T= 100^{\circ}C$

(b)

Now, the average rate of change of the temperature, with respect to distance, from the beginning of the trip to the end of the trip be calculated as:

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$y=m.x+c$

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(d)

comparing the given function of the temperature with the general equation of a straight line:

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Part b)

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