Answered

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(a) A simple pendulum oscillates back and forth on a space vehicle. An astronaut on the space vehicle measures the period of the pendulum to be 22.58 seconds (it is a big pendulum). A passing observer in another space ship measures the period to be 31.87 seconds. Determine the relative velocity between the two observers. Show all calculation steps.​

Sagot :

oladayoademosu

Answer:

0.706

Explanation:

Since the other astronaut measures a longer time, this is a time dilation problem. So, our equation for time dilation is given by

T = T₀/√(1 - β²) where T = period on passing space ship = 31.87 s, T₀ = period on other space vehicle = proper time = 22.58 s and β = relative velocity of between the two observers.

T = T₀/√(1 - β²)

√(1 - β²) = T₀/T

squaring both sides, we have

[√(1 - β²)]² = (T₀/T)²

1 - β² = (T₀/T)²

β² = 1 - (T₀/T)²

taking square root of both sides, we have

√β² = √[1 - (T₀/T)²]

β = √[1 - (T₀/T)²]

substituting the values of the variables into the equation, we have

β = √[1 - (22.58 s/31.87 s)²]

β = √[1 - (0.7085)²]

β = √[1 - 0.502]

β = √0.498

β = 0.706

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