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3 years ago

What is the gravitational potential energy of a 3 kg ball kicked into the air at a height of 5 meters?

Physics

1 answer:

sladkih [1.3K]3 years ago

7 0

formula for gravitational P.E =mgh

Solution:-mass=3kg height=5metre and gravity=9.8 or 10m/sec² so P.E=mgh , 3×9.8×5=147kgm²/sec²

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3 years ago

Question :

irakobra [83]

A rubber ball and a stone of the same size are examples which will have more inertia and is therefore denoted as option A.

<h3>What is Inertia?</h3>

This is referred to as the property exhibited by a body in which it has the tendency to remain at rest or in uniform motion.This property is dependent on the mass of the substance as we can deduce that the greater the mass, the greater the inertia and vice versa.

The size of a rubber ball and stone will have different masses in which that of the stone will be greater. This is as a result of the difference in the nature of the substances which are used to make both items mentioned above.

This is therefore the reason why a rubber ball and a stone of the same size as having more inertia(mass) where chosen as the most appropriate choice in this scenario.

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2 years ago

A company records authors reading their books aloud. These recordings are shared with libraries around the world for people to l

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2 years ago

Read 2 more answers

The ratio of Earth's radius to that of Mars (RR/RM) is approximately 19/10, and the ratio of the density of Earth to that of Mar

drek231 [11]

Answer:

The answer is letter c.

Explanation:

We need to find the relation of masses first and then apply the equivalence principle.

We know that the ratio of the density of Earth to that of Mars is approximately 14/10.

$\frac{\rho_{E}}{\rho_{M}}=\frac{14}{10}$

Density is the mass divided by the volume (Here the volume of each planet is a solid sphere of uniform density), so

$\rho_{E}=\frac{M_{E}}{V_{E}}=\frac{M_{E}}{\frac{4}{3}\pi R_{E}^{3}}$

$\rho_{M}=\frac{M_{M}}{V_{M}}=\frac{M_{E}}{\frac{4}{3}\pi R_{M}^{3}}$

Then, the ratio between the densities will be:

$\frac{\rho_{E}}{\rho_{M}}=\frac{M_{E}R_{M}^{3}}{M_{M}R_{E}^{3}}=\frac{14}{10}$

We know that $\frac{R_{E}}{R_{M}}=19/10$ , then we can write the above equation as:

$\frac{M_{E}}{M_{M}}=\left(\frac{R_{E}}{R_{M}}\right)^{3} \frac{14}{10}$

$\frac{M_{E}}{M_{M}}=(\frac{19}{10})^{3} \frac{14}{10}$

Now, using the gravitational force:

To the earth

$g=G\frac{M_{E}}{R_{E}^{2}}$

To the mars

$g=G\frac{M_{M}}{R_{M}^{2}}$

But, we know that $M_{M}=(\frac{10}{19})^{3} \frac{10}{14}M_{E} \: and \: R_{M}=\frac{10}{19}R_{E}$

Therefore, we will have:

$g=G\frac{(\frac{10}{19})^{3} \frac{10}{14}M_{E}}{(\frac{10}{19}R_{E})^{2}}$

$g=\frac{10*10}{14*19}G\frac{M_{E}}{R_{E}^{2}}$

$g_{M}=0.38g$

$g_{M} \approx \frac{2}{5}g$

The answer is letter c.

I hope it helps you!

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3 years ago

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Answer:

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Explanation:

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3 years ago