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weeeeeb [17]
2 years ago
10

What energy transformation occurs when a skydiver first jump off a plane

Physics
1 answer:
Rus_ich [418]2 years ago
5 0
Hello,

When a skydiver first jumps off a plane potential energy (<span>energy in an object due to its position) will be converted into kinetic energy (</span><span>energy which a body possesses because of its motion.)
</span>
Faith xoxo


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An astronomer whose secret hobby is riding merry-go-rounds has dedicated his career to finding the stars that rotate the most ra
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An astronomer whose secret hobby is riding merry-go-rounds has dedicated his career to finding the stars that rotate the most rapidly. But the stars are all very far and hence he can not identify the starts that spins more rapidly.

<u>Explanation:</u>

The things in the world rotates. the earth, sun star everything will be spinning. There is a star named Achenar. this is the start that spins at a faster rate. It is the biggest star that stays at the position 10. Its mass is 7 times greater than the sun's mass.The shining of the star can be measured by luminosity. It is the measure of the brightness of the star.

The energy that is given by a star in a second of time is the luminosity of the star. All stars are present in the different distances form the surface of the earth. but when we look from the earth or even when when go at certain altitude we can find all stars to be located at the same distance. hence if we need to find the star that spins more rapidly, then the star spectral lines can be observed. The stars that rotates will be having the wide lines in the spectra.

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3 years ago
Three resistors are connected into the section of a circuit described by the diagram. At which labeled point or points of the ci
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Answer: Point Z only

Explanation: Just took the quiz

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What can you calculate using the equation P equals W/t​
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WE CALCULATE POWER AND RATE OF DOING WORK IS CALLED POWER

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Is there a frame of reference one can go into that seems to eliminate gravity as Newton described it?
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3 years ago
Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o
Pachacha [2.7K]

(a) 0.448

The gravitational potential energy of a satellite in orbit is given by:

U=-\frac{GMm}{r}

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance of the satellite from the Earth's centre, which is sum of the Earth's radius (R) and the altitude of the satellite (h):

r = R + h

We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as

\frac{U_B}{U_A}=\frac{-\frac{GMm}{R+h_B}}{-\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}

and so, substituting:

R=6370 km\\h_A = 5970 km\\h_B = 21200 km

We find

\frac{U_B}{U_A}=\frac{6370 km+5970 km}{6370 km+21200 km}=0.448

(b) 0.448

The kinetic energy of a satellite in orbit around the Earth is given by

K=\frac{1}{2}\frac{GMm}{r}

So, the ratio between the two kinetic energies is

\frac{K_B}{K_A}=\frac{\frac{1}{2}\frac{GMm}{R+h_B}}{\frac{1}{2}\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}

Which is exactly identical to the ratio of the potential energies. Therefore, this ratio is also equal to 0.448.

(c) B

The total energy of a satellite is given by the sum of the potential energy and the kinetic energy:

E=U+K=-\frac{GMm}{R+h}+\frac{1}{2}\frac{GMm}{R+h}=-\frac{1}{2}\frac{GMm}{R+h}

For satellite A, we have

E_A=-\frac{1}{2}\frac{GMm}{R+h_A}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+5.97\cdot 10^6 m}=-4.65\cdot 10^8 J

For satellite B, we have

E_B=-\frac{1}{2}\frac{GMm}{R+h_B}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+21.2\cdot 10^6 m}=-2.08\cdot 10^8 J

So, satellite B has the greater total energy (since the energy is negative).

(d) -2.57\cdot 10^8 J

The difference between the energy of the two satellites is:

E_B-E_A=-2.08\cdot 10^8 J-(-4.65\cdot 10^8 J)=-2.57\cdot 10^8 J

4 0
3 years ago
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