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satellite originally moves in a circular orbit of radius R around the Earth. Suppose it is moved into a circular orbit of radius 4R.
(i) What does the force exerted on the satellite then become?
eight times larger<span>four times larger </span>one-half as largeone-eighth as largeone-sixteenth as large(ii) What happens to the satellite's speed?<span>eight times larger<span>four times larger </span>one-half as largeone-eighth as largeone-sixteenth as large(iii) What happens to its period?<span>eight times larger<span>four times larger </span>one-half as largeone-eighth as largeone-sixteenth as large</span></span>
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Mountains, tops of buildings, and high-flying aircraft are all part of Earth's atmosphere, no matter how high they are. On the other hand, space doesn't belong to our atmosphere, it is outside of it. Having this in mind, the best location to place a telescope used to observe x-rays from stars is in space.
Answers are:
(1) KE = 1 kg m^2/s^2
(2) KE = 2 kg m^2/s^2
(3) KE = 3 kg m^2/s^2
(4) KE = 4 kg m^2/s^2
Explanation:
(1) Given mass = 0.125 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.125 * (16)
KE = 1 kg m^2/s^2
(2) Given mass = 0.250 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.250 * (16)
KE = 2 kg m^2/s^2
(3) Given mass = 0.375 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.375 * (16)
KE = 3 kg m^2/s^2
(4) Given mass = 0.500 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.5 * (16)
KE = 4 kg m^2/s^2
False, as an object falls its potential energy turns into kinetic energy thus decreasing the potential energy.
