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laila [671]
3 years ago
13

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 3.17 s for a ball to fall freely 11.26 m, starting from rest. What is the acceleration due to gravity on that planet?
Physics
1 answer:
devlian [24]3 years ago
7 0

Answer:

2.24 m/s²

Explanation:

Using equation of motion

s = ut + \frac{1}{2}at²

u = 0 , t = 3.17 s , s = 11.26 m

Put these values in the equation above

11.26 = 0 +.5 x a( 3.17)²

a = 2.24 ms⁻².

So acceleration due to gravity on that planet will be 2.24 m s⁻².

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Two point charges are separated by 10 cm, with an attractive force between them of 15 N. Find the force between them when they a
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Answer:

(a) the force is 8.876 N

(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

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when the distance between the charges changes to 13 cm (0.13 m)

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Apply Coulomb's law;

F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N

Part (b)

the magnitude of each charge, if they have equal magnitude

F = \frac{KQ^2}{r^2}

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F is the force between the charges

K is Coulomb's constant

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r is the distance between the charges

F = \frac{KQ^2}{r^2} \\\\Q = \sqrt{\frac{Fr^2}{K} } \\\\Q =  \sqrt{\frac{15*(0.1)^2}{8.99*10^9} } = 4.085 *10^{-6} \ C\\\\Q = 4.085 \ \mu C

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