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vodomira [7]
3 years ago
14

The gravitational acceleration on Mars is 3.71 m/s2. If the pendulums were set in motion on the red planet, how would that affec

t their periods?
Physics
1 answer:
Elanso [62]3 years ago
6 0
On Earth, the period of a pendulum is given by:
T_{earth}=2\pi  \sqrt{ \frac{L}{g_{earth} }
where L is the length of the pendulum and g_{earth}=9.81~m/s^2 is the gravitational acceleration on Earth.
Similarly, the period of the same pendulum on Mars will be
T_{mars}=2\pi \sqrt{ \frac{L}{g_{mars} }
where g_{mars}=3.71~m/s^2 is the gravitational acceleration on Mars.
Therefore, if we want to see how does the period of the pendulum on Mars change compared to the one on Earth, we can do the ratio between the two of them:
\frac{T_{mars}}{T_{earth}}= \sqrt{ \frac{g_{earth}}{g_{mars}} }  =  \sqrt{ \frac{9.81~m/s^s}{3.71~m/s^2} }=1.63
Therefore, the period of the pendulum on Mars will be 1.63 times the period on Earth.
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Answer:

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Explanation:

Given that

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Light of wavelength 560 nm passes through a slit of width 0. 170 mm. (a) the width of the central maximum on a screen is 8. 00 m
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The distance between slit and the screen is 1.214m.

To find the answer, we have to know about the width of the central maximum.

<h3>How to find the distance between slit and the screen?</h3>
  • It is given that, wavelength 560 nm passes through a slit of width 0. 170 mm, and the width of the central maximum on a screen is 8. 00 mm.
  • We have the expression for slit width w as,

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where, d is the distance between slit and the screen, and a is the slit width.

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Thus, we can conclude that, the distance between slit and the screen is 1.214m.

Learn more about the width of the central maximum here:

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The Answer is B

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