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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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dolphi86 [110]

Answer:

Explanation:

To convert gram / centimeter³ to kg / m³

gram / centimeter³

= 10⁻³ kg / centimeter³

= 10⁻³  / (10⁻²)³ kg / m³

= 10⁻³ / 10⁻⁶ kg / m³

= 10⁻³⁺⁶ kg / m³

= 10³ kg / m³

So we shall have to multiply be 10³ with amount in gm / cm³ to convert it into kg/m³

2.33 gram / cm³

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3 0
2 years ago
You are on vacation in San Francisco and decide to take a cable car to see the city. A 5800-kgkg cable car goes 260 mm up a hill
Stella [2.4K]

Answer:

4.325\times10^6J

Explanation:

Mass of the cable car, m = 5800 kg

It goes 260 m up a hill, along a slope of \theta=17^o

Therefore vertical elevation of the car = 260sin\theta=260sin17^o=76.0166m

Now, when you get into the cable car, it's velocity is zero, that is, initial kinetic energy is zero (since K.E. = \frac{1}{2} mv^2). Similarly as the car reaches the top, it halts and hence final kinetic energy is zero.

Therefore the only possible change in the cable car system is the change in it's gravitational potential energy.

Hence, total change in energy = mgh = 5800\times9.81\times76.0166J=4.325\times10^6J

where, g = acceleration due to gravity

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4 0
2 years ago
A 0.305 kg book rests at an angle against one side of a bookshelf. The magnitude and direction of the total force exerted on the
tankabanditka [31]

Answer

given,

F_L= 1.52\ N

\theta_L= 31^0

mass of book = 0.305 Kg

so, from the diagram attached  below

F_L cos {\theta_L} + F_b sin {\theta_b} = m g

1.52 times cos {31^0} + F_b sin {\theta_b} = 0.305 \times 9.8

F_b sin {\theta_b} = 2.989 -1.303

F_b sin {\theta_b} = 1.686

computing horizontal component

F_b cos {\theta_b} = F_L sin {\theta_L}

cos {\theta_b} = \dfrac{F_L sin {\theta_L}}{F_b}

cos {\theta_b} = \dfrac{1.52 \times sin {31^0}}{1.686}

cos {\theta_b} = 0.464

θ = 62.35°

5 0
2 years ago
The temperature of a heat engine is 500k some of the heat generated by the engine flows to the surroundings which are at a temp
Ierofanga [76]

1-125/500)x100=efficiency

1-1/4)x100 =75pc

3 0
1 year ago
Soot particles ("black carbon aerosols" generally cause ________ of earth's atmosphere by ________ solar energy.
sineoko [7]
They cause an increase in temp of earths atmosphere or warming by absorbing solar energy. hope this helps
7 0
1 year ago
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