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ValentinkaMS [17]
3 years ago
10

A spaceship travels toward the Earth at a speed of 0.97c. The occupants of the ship are standing with their torsos parallel to t

he direction of travel. According to Earth observers, they are about 0.50 m tall and 0.50 m wide. Calculate what the occupants’ height and width according to the others on the spaceship?
Physics
1 answer:
Ket [755]3 years ago
3 0

Answer:

Explanation:

We shall apply length contraction einstein's relativistic formula to calculate the length observed by observer on the earth . For the observer , increased length will be observed for an observer on the earth

L=\frac{.5}{\sqrt{1-(\frac{.97c}{c})^2 } }

L=\frac{.5}{.24}

L= 2.05

The length will appear to be 2.05 m . and width will appear to be .5 m  to the observer on the spaceship. . It is so because it is length which is moving parallel to the direction of travel. Width will remain unchanged.  

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Answer:

The value is the temperature of the air inside the tire T_{2} = 340.54 K

% of the original mass of air in the tire should be released 99.706 %

Explanation:

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(a) Volume of the system is constant so  pressure is directly proportional to the temperature.

\frac{T_{2} }{T_{1} } = \frac{P_{2} }{P_{1} }

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This is the value is the temperature of the air inside the tire

(b). Since volume of the tyre is constant & pressure reaches the original value.

From ideal gas equation P V = m R T

Since P , V & R is constant. So

m T = constant

m_{1}  T_{1} =  m_{2}  T_{2}

\frac{m_{2} }{m_{1}  } = \frac{T_{1} }{T_{2} }

\frac{m_{2} }{m_{1}  } = \frac{300}{354.54}

\frac{m_{2} }{m_{1}  } =0.00294

value of  the original mass of air in the tire should be released is  \frac{m_{2} - m_{1}}{m_{1}}

\frac{0.00294-1}{1}

⇒ -0.99706

% of the original mass of air in the tire should be released 99.706 %.

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