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An astronaut moving with a speed of 0.84c relative to Earth measures her heart rate to be 71 beats per minute.

**Part A**
When an Earth-based observer measures the astronaut's heart rate, is the result:
A. less than 71 beats per minute
B. greater than 71 beats per minute
C. equal to 71 beats per minute

**Part B**
Explain your answer.
(Essay answers are limited to about 500 words or 3800 characters, including spaces.)

Calculate the astronaut's heart rate as measured on Earth.
Express your answer using two significant figures.

f = _____ beats/min

Answer :

Part A: Earth-based observer measures the astronaut's heart rate, it will be greater than 71 beats per minute due to the time dilation effect caused by the astronaut's high speed. The correct answer is option B.

The calculated heart rate as measured on Earth is approximately 131 beats per minute.

The astronaut is moving with a speed of 0.84c relative to Earth and measures her heart rate to be 71 beats per minute.

Part A: When an Earth-based observer measures the astronaut's heart rate, is the result greater than, less than, or equal to 71 beats per minute?

The heart rate measured by an Earth-based observer will be greater than 71 beats per minute. This is because of the time dilation effect predicted by Einstein's theory of relativity. When an object is moving at high speeds relative to an observer, time slows down for that object. This means that the astronaut's heart rate appears slower to her, but to an observer on Earth, it will appear faster.

Part B: Calculate the astronaut's heart rate as measured on Earth.

To calculate the astronaut's heart rate as measured on Earth, we can use the concept of time dilation. The formula for time dilation is:

Δt' = Δt / √(1 - (v^2/c^2))

Where:
Δt' is the time measured by the Earth-based observer
Δt is the time measured by the astronaut
v is the relative velocity of the astronaut (0.84c)
c is the speed of light in a vacuum (3.00 x 10^8 m/s)

Let's substitute the values into the formula:

Δt' = 71 / √(1 - (0.84c)^2/c^2)

Simplifying the equation:

Δt' = 71 / √(1 - 0.84^2)

Δt' = 71 / √(1 - 0.7056)

Δt' = 71 / √(0.2944)

Δt' = 71 / 0.542

Δt' ≈ 131 beats per minute

Therefore, the astronaut's heart rate, as measured by an Earth-based observer, is approximately 131 beats per min.

To learn more about speed click here:

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