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3 years ago

Which has more momentum, a speeding baseball or an ocean liner at rest in a harbor?

1 answer:

4 0

Momentum is (mass) times (speed), so nothing that is at rest has any momentum. If the battleship is at rest, then a mosquito in flight, a leaf falling from a tree, and your speedy baseball each have more momentum than the ship has.

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-x times -x what is the answer

Brilliant_brown [7]

Answer:

Explanation:

5 0

2 years ago

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Atmospheric pressure is greater at the base of a mountain than at

aliina [53]

At the top of the mountain, when he tightens the cap onto the bottole, there is some water and some air inside the bottle. Then he brings the bottle down to the base of the mountain.

The pressure on the outside of the bottle is greater than it was when he put the cap on. If anything could get out of the bottlde, it would. But it can't . . . the cap is on too tight. So all the water and all the air has to stay inside, and anything that can get squished into a smaller space has to get squished into a smaller space.

The water is pretty much unsquishable.

Biut the air in there can be <em>COMPRESSED</em>. The air gets squished into a smaller space, and the bottle wrinkles in slightly.

8 0

2 years ago

What is meant by 'wasted energy'?

schepotkina [342]

Wasted energy is an energy which is transformed without significant use.

5 0

3 years ago

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A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o

blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

$\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$T$ - Period of rotation of the satellite, measured in seconds.

$r$ - Distance of the satellite with respect to the center of the planet, measured in meters.

$G$ - Gravitational constant, measured in newton-square meters per square kilogram.

$M$ - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

$r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}$

$r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} }$ (2)

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ and $T = 25800\,s$ , then the distance of the satellite is:

$r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }$

$r \approx 18.897\times 10^{6}\,m$

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

$F = \frac{G\cdot m\cdot M}{r^{2}}$ (3)

Where:

$m$ - Mass of the satellite, measured in kilograms.

$F$ - Force exerted on the satellite by the Earth, measured in newtons.

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ , $m = 6105\,kg$ and $r \approx 18.897\times 10^{6}\,m$ , then the gravitational force is: