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3 years ago

Describe how you could use two photographs taken at a different time to prove that the moon is in motion

Physics

2 answers:

Alla [95]3 years ago

7 0

Answer:

There are following methods by which you describe that moon is in motion with the help of two photographs.

Explanation:

First you can take a photo on one night at a particular time, at a particular angle which you fixed and at a particular location now take another photo in the next night at different time, at same location and at same angle. Then you observe that the moon is in motion.

Next you can take two photographs in one night span, in this method you can take photo of moon at 10'o clock in the night and observe the stars around it then you can take another photo after 2 hours and observe the stars around it. Then you will find that the position of the moon is shifted or arrangement of the stars is not same in two cases then you can says that moon is in motion.

Slav-nsk [51]3 years ago

5 0

You can prove that moon is in motion by taking two photos with the following features:

They are taken at the same hour and minute.
They are taken on two different nights.
They are taken from the same location.

This would allow an accurate comparison and would should the moon's motion.

I hope this helped! :)

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3 years ago

Read 2 more answers

Suppose the earth is shaped as a sphere with radius 4,0004,000 miles and suppose it rotates once every 24 hours. How many miles

Alenkinab [10]

Answer:

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$L' = \frac{25133 mi}{24h}=1047 mi$

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2 years ago

A football is kicked from the ground with a velocity of 38m/s at an angle of 40 degrees and eventually lands at the same height.

Anastasy [175]

Initially, the velocity vector is $\langle 38cos(40^{\circ}),38sin(40^{\circ}) \rangle=\langle 29.110, 24.426 \rangle$ . At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by $4.9(0.2)^2$ , so the velocity is $\langle 29.110, 24.426-0.196 \rangle = \langle 29.110, 24.23 \rangle$ .

Converting back to direction and magnitude, we get $\langle r,\theta \rangle=\langle \sqrt{29.11^2+24.23^2},tan^{-1}(\frac{29.11}{24.23}) \rangle = \langle 37.87,50.2^{\circ}\rangle$

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