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STatiana [176]
2 years ago
4

Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 10,000 kg. The thrust of its engines is

30,000 N. (a) Calculate its the magnitude of acceleration in a vertical takeoff from the Moon. (b) Could it lift off from Earth? If not, why not? If it could, calculate the magnitude of its acceleration.
Physics
1 answer:
Ratling [72]2 years ago
5 0

Answer:

Part a)

a = 1.37 m/s/s

Part b)

since the force of gravity is more than the thrust force of rocket so it will not lift off the surface of Earth

Explanation:

Part a)

Net force on the Module due to thrust of engine is given as

F = 30,000 N

now net force while is ejected out of the surface of moon is given as

F_{net} = F - F_g

here we know that

F_g = mg_{moon}

where we have

g_{moon} = \frac{g}{6}

F_{net} = 30,000 - (10000)(\frac{9.8}{6})

F_{net} = 13666.67 N

now the acceleration of the module on the moon is

a = \frac{F_{net}}{m}

a = \frac{13666.67}{10,000} = 1.37 m/s^2

Part b)

Now on the surface of earth the force of gravity on the module is given as

F_g = mg

F_g = 10,000 \times 9.8

F_g = 98000 N

since the force of gravity is more than the thrust force of rocket so it will not lift off the surface of Earth

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An ideal gas is contained in a vessel at 300 K. The temperature of the gas is then increased to 900 K. (i) By what factor does t
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The question is missing some parts. Here is the complete question.

An ideal gas is contained in a vessel at 300K. The temperature of the gas is then increased to 900K.

(i) By what factor does the average kinetic energy of the molecules change, (a) a factor of 9, (b) a factor of 3, (c) a factor of \sqrt{3}, (d) a factor of 1, or (e) a factor of \frac{1}{3}?

Using the same choices in part (i), by what factor does each of the following change: (ii) the rms molecular speed of the molecules, (iii) the average momentum change that one molecule undergoes in a colision with one particular wall, (iv) the rate of collisions of molecules with walls, and (v) the pressure of the gas.

Answer: (i) (b) a factor of 3;

              (ii) (c) a factor of \sqrt{3};

              (iii) (c) a factor of \sqrt{3};

             (iv) (c) a factor of \sqrt{3};

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KE=\frac{3}{2}nRT

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n is mols

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For temperature and energy, the factor of change is 3.

(ii) Rms is root mean square velocity and is defined as

V_{rms}=\sqrt{\frac{3k_{B}T}{m} }

Calculating velocity for each temperature:

For 300K:

V_{rms1}=\sqrt{\frac{3k_{B}300}{m} }

V_{rms1}=30\sqrt{\frac{k_{B}}{m} }

For 900K:

V_{rms2}=\sqrt{\frac{3k_{B}900}{m} }

V_{rms2}=30\sqrt{3}\sqrt{\frac{k_{B}}{m} }

Comparing both veolcities:

\frac{V_{rms2}}{V_{rms1}}= (30\sqrt{3}\sqrt{\frac{k_{B}}{m} }) .\frac{1}{30} \sqrt{\frac{m}{k_{B}} }

\frac{V_{rms2}}{V_{rms1}}=\sqrt{3}

For rms, factor of change is \sqrt{3}

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q = m.v

Since velocity has a factor of \sqrt{3} and velocity and momentum are proportional, average momentum change increase by a factor of

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</span>

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Answer:

33.33 %

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7 0
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