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svp [43]
2 years ago
14

An astronaut takes what he measures to be a 10-min nap in a space station orbiting Earth at 8000 m/s. A signal is sent from the

station to Earth at the instant he falls asleep, and a second signal is sent to Earth the instant he wakes. Part A Does an observer on Earth measure the astronaut's nap as being shorter than, longer than, or equal to 10 min? Does an observer on Earth measure the astronaut's nap as being shorter than, longer than, or equal to 10 ? a. equal b. shorter c. longer
Physics
1 answer:
svet-max [94.6K]2 years ago
3 0

Answer:

longer than

Explanation:

given,

time of nap = 10 min

speed of orbiting earth = 8000 m/s

c is the speed of light

using the equation of time dilation

t' = \dfrac{t}{\sqrt{1-\dfrac{v^2}{c^2}}}

now inserting all the values

t' = \dfrac{10}{\sqrt{1-\dfrac{8000^2}{3\times 10^8)^2}}}

t' = \dfrac{10}{0.9999}

t' = 10.001 s

on solving the above equation we will get a value greater than 10minutes.

hence, On earth time of nap measured will be longer than 10 min

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How high above the earth's surface is g reduced to 8.80m/^2?
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Gravity changes as the altitude change.<span> The gravitational force is proportional to 1/R2,  where R is the distance from the center of the Earth the radius of earth where gravity is 9.8 m/s^2 is 6400 km this will serve as the zero mark.

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6 0
2 years ago
Read 2 more answers
An engineer uses aluminum to build an airplane rather than composite materials that are lighter and stronger. He does this becau
AleksandrR [38]

Answer:

choosing a material that will show warning before it fails

Explanation:

According to my research on different architectural engineering techniques, I can say that based on the information provided within the question this is an example of choosing a material that will show warning before it fails. By choosing aluminum he can detect certain failures a long time before it actually happens since aluminum shows signs of wear and tear and doesn't just break immediately.

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2 years ago
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In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that
mars1129 [50]

Answer:

The maximum transverse speed of the bead is 0.4 m/s

Explanation:

As we know that the Amplitude of the travelling wave is

A = 3.65 mm

Now the speed of the travelling wave is

v_x = 13.5 m/s

now we know that distance of first antinode from one end is 27.5 cm

so length of the loop of the standing wave is given as

\frac{\lambda}{4} = 27.5 cm

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now we have

N = \frac{2L}{\lambda}

N = \frac{2(1.65)}{1.10}

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now we have

R = 2A sin(kx)

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now we have

f = \frac{v}{\lambda}

f = \frac{13.5}{1.1}

f = 12.27 Hz

now maximum speed is given as

v_y = R\omega

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v_y = 0.4 m/s

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