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2 years ago

the speed of travel of the moon around the earth, using the formula for the speed of a moving object in a circular path

Physics

1 answer:

Svetach [21]2 years ago

5 0

Answer: 1018.26 m/s

Explanation:

Approaching the orbit of the Moon around the Earth to a circular orbit (or circular path), we can use the equation of the speed of an object with uniform circular motion:

$V=\sqrt{G\frac{M}{r}}$

Where:

$V$ is the speed of travel of the Moon around the Earth

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=5.972(10)^{24} kg$ is the mass of the Earth

$r=384400(10)^{3} m$ is the distance from the center of the Earth to the center of the Moon

Solving:

$V=\sqrt{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{5.972(10)^{24} kg}{384400(10)^{3} m}}$

$V=1018.26 m/s$ This is the speed of travel of the Moon around the Earth

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Answer: MR²

is the the moment of inertia of a hoop of radius R and mass M with respect to an axis perpendicular to the hoop and passing through its center

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Since in the hoop , all mass elements are situated at the same distance from the centre , the following expression for the moment of inertia can be written as follows.

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AlekseyPX

Answer:

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Explanation:

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A car initially traveling at 24 m/s slams on the brakes and moves forward 196 m before coming to a complete halt. What was the m

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Answer:

$-1.47 m/s^2$

Explanation:

We can use the following SUVAT equation to solve the problem:

$v^2 - u^2 = 2ad$

where

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u = 24 m/s is the initial velocity

a is the acceleration

d = 196 m is the displacement of the car before coming to a stop

Solving the equation for a, we find the acceleration:

$a=\frac{v^2-u^2}{2d}=\frac{0-(24)^2}{2(196)}=-1.47 m/s^2$

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