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boyakko [2]
3 years ago
13

A man on the Moon observes two spaceships coming toward him from opposite directions at speeds of 0.600c and 0.600c. What is the

relative speed of the two ships as measured by a passenger on either one of the spaceships
Physics
1 answer:
vaieri [72.5K]3 years ago
5 0

Answer:

If we use the equation for the transformation of velocities for moving frames:

v' = (v - u) / (1 - u * v / c^2) where we measure the speed of v' approaching from the left where v is in a frame moving at -u towards v'

v' = (.6 c - (-.6 c)) / (1 - (-.6 c) * .6 c / c^2) = 1.2 c / (1 + .6 * .6)

or v' = 1.2 c / (1 + .36) = .88 c

v is approaching from the left at .6 c in the reference frame and the other frame approaches from the right at -.6 c with speed u  (-.6 c) and we measure the speed of v as seen in the frame moving to the left

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The Earth is a sphere is a statement to describe a scientific law

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The line of latitude that is zero degrees latitude is the
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Equator

Explanation:

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Twenty is the _________________ of potassium
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Choose the best explanation from among the following: A. It takes more energy to go from the Moon to the Earth because the Moon
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B. The escape speed of the Moon is less than that of the Earth; therefore, less energy is required to leave the Moon.

Explanation:

Since the speed required to escape from the gravitational attraction of the Moon is less than the speed required to escape from the gravitational attraction of the Earth, less energy is required to travel from the Moon to the Earth, than is required to travel from the Earth to the Moon. This is because the kinetic energy is directly proportional to the square of the velocity.

8 0
3 years ago
Very large forces are produced in joints when a person jumps from some height to the ground. Calculate the force produced if an
krok68 [10]

Answer:

A) 31 kJ

B)  1.92 KJ

C) 40 ,  2.48

Explanation:

weight of person ( m ) = 79 kg

height of jump ( h ) = 0.510 m

Compression of joint material ( d ) = 1.30 cm ≈  0.013 m

A) calculate the force

Fd = mgh

F = mgh / d

W = mg

F(net) = W + F  = mg ( 1 + \frac{h}{d} )

          =  79 * 9.81 ( 1 + (0.51 / 0.013) )

          = 774.99 ( 40.231 ) ≈ 31 KJ

B) calculate the force when the stopping distance = 0.345 m

d = 0.345 m

Fd = mgh  hence  F = mgh / d

F(net) = W + F  = mg ( 1 + \frac{h}{d} )

          =  79 * 9.81 ( 1 + (0.51 / 0.345) )

          = 774.99 ( 2.478 ) = 1.92 KJ

C) Ratio of force in part a with weight of person

=  31000 / ( 79 * 9.81 ) = 31000 / 774.99 = 40

  Ratio of force in part b with weight of person

= 1920 / 774.99 = 2.48

4 0
3 years ago
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