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2 years ago

A cylindrical space colony 8 km in diameter and 30 km long has been proposed as living quarters for future space explorers.

Physics

1 answer:

Dafna1 [17]2 years ago

4 0

Answer:

$\omega=0.05rad/s$

Explanation:

The gravitational field (fictitious acceleration) produced would be equal to the centripetal acceleration needed to move as a circle (acceleration of the system). The equation for centripetal acceleration is $a_{cp}=r\omega^2$ , which means that the angular velocity of the cylindrical space colony, which has a radius r=4km and needs to have a gravitational field of 1-g ( $9.8m/s^2$ ) needs to be:

$\omega=\sqrt{\frac{a_{cp}}{r}}=\sqrt{\frac{9.8m/s^2}{4000m}}=0.05rad/s$

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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

Dahasolnce [82]

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 $\sqrt{3}$ ;

(v) (e) a factor of 3;

Explanation: (i) Kinetic energy for ideal gas is calculated as:

$KE=\frac{3}{2}nRT$

where

n is mols

R is constant of gas

T is temperature in Kelvin

As you can see, kinetic energy and temperature are directly proportional: when tem perature increases, so does energy.

So, as temperature of an ideal gas increased 3 times, kinetic energy will increase 3 times.

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 $\sqrt{3}$

(iii) Average momentum change of molecule depends upon velocity:

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

(iv) Collisions increase with increase in velocity, which increases with increase of temperature. So, rate of collisions also increase by a factor of $\sqrt{3}$ .

(v) According to the Pressure-Temperature Law, also known as Gay-Lussac's Law, when the volume of an ideal gas is kept constant, pressure and temperature are directly proportional. So, when temperature increases by a factor of 3, Pressure also increases by a factor of 3.

4 0

2 years ago

A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius 2.0 m). When the mass is at the highest point

il63 [147K]

Answer:

31.1 N

Explanation:

m = mass attached to string = 0.50 kg

r = radius of the vertical circle = 2.0 m

v = speed of the mass at the highest point = 12 m/s

T = force of the string on the mass attached.

At the highest point, force equation is given as