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4 years ago

10

An ideal monatomic gas at 275 K expands adiabatically and reversibly to six times its volume. What is its final temperature (in

K)

1 answer:

Gwar [14]4 years ago

3 0

The final temperature is 83 K.

<u>Explanation</u>:

For an adiabatic process,

$T {V}^{\gamma - 1} = \text{constant}$

$\cfrac{{T}_{2}}{{T}_{1}} = {\left( \cfrac{{V}_{1}}{{V}_{2}} \right)}^{\gamma - 1}$

Given:-

${T}_{1} = 275 \; K$

${T}_{2} = T \left( \text{say} \right)$

${V}_{1} = V$

${V}_{2} = 6V$

$\gamma = \cfrac{5}{3} \;$ (the gas is monoatomic)

$\therefore \cfrac{T}{275} = {\left( \cfrac{V}{6V} \right)}^{\frac{5}{3} - 1}$

$\Rightarrow \cfrac{T}{275} = {\left( \cfrac{1}{6} \right)}^{\frac{2}{3}}$

T = 275 $\times$ 0.30

T = 83 K.

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An automobile travels on a straight road for 40 km at 30 km/h. It then continues in the same direction for another 40 km at 60 k

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Answer:

The average velocity is 40km/h.

Explanation:

The average velocity is $\bar{v}=\frac{\Delta x }{\Delta t}$ , where $\Delta x$ is the distance traveled and $\Delta t$ the time elapsed.

The distance traveled is clearly 80km since it's all done in the same direction, we only need to know the time elapsed. For this we calculate the time elapsed on the first part, and add it to the time elapsed on the second part using always the formula $\Delta t=\frac{\Delta x }{v}$ , where v is the velocity on each part, which is constant.

The time elapsed for the first part is $\Delta t_1=\frac{40 km}{30km/h}=\frac{4}{3}h$ , and the time elapsed for the second part is $\Delta t_2=\frac{40 km}{60km/h}=\frac{2}{3}h$ , giving us a total time of $\Delta t_1+\Delta t_2=\frac{4}{3}h+\frac{2}{3}h$ =2h.

Finally, we can calculate the average velocity: $\bar{v}=\frac{80km}{2h}=40km/h$ .

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3 years ago

Numerical Problems

Evgesh-ka [11]

Answer:

a = 5 [m/s²]

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f}=v_{o}+a*t$

where:

Vf = final velocity = 20 [m/s]

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3 years ago

A 405 Hz tuning fork and a piano key are struck together, and no beats are heard. When a 402 Hz tuning fork and the same piano k

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The difference between the frequencies of the piano key and the tuning fork gives the frequency of the beats.

When the tuning fork is 405 Hz, and no beats are heard, then the piano key is also 405 Hz.

When the piano key is 405 Hz and the tuning fork is 402 Hz, then 405 - 402 = 3 beats are heard.

The piano key is 405 Hz.

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4 years ago

If gravity on the earth increased, what affect would it have on the moon

Rufina [12.5K]

Answer:

If gravity on Earth is increased, this gravitational tugging would have influenced the moon's rotation rate. If it was spinning more than once per orbit, Earth would pull at a slight angle against the moon's direction of rotation, slowing its spin. If the moon was spinning less than once per orbit, Earth would have pulled the other way, speeding its rotation.

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3 years ago

At the same instant that a 0.50 kg ball is dropped from 25 m above Earth, a second ball, with a mass of .25 kg, is thrown straig

Anettt [7]

Answer:

A. 7.1m

B. 3.55m/s

C. 1.775m/s^2

Explanation:

First step is to identify given parameters;

Ball 1: m₁ = 0.5kg, u (initial velocity) =0, t = 2seconds

Ball 2: m₂ = 0.25kg, u = 15m/s, t = 2seconds

<u>Second step:</u> we determine the y-coordinate of ball 1 after 2 seconds, using the equation of motion under gravity as shown below;

$y = ut - \frac{gt^2}{2}$

$y_{1} = 0 X 2 - \frac{9.8 X2^2}{2}$

$y_{1} = -19.6m$

Recall, that the ball was thrown from a height of 25m, total y-coordinate of ball 1 after 2 seconds becomes 25m +(-19.6m)

[tex]y_{1} = 5.4m[/tex]

<u>Third step</u>: we determine the y-coordinate of ball 2 after 2 seconds

$y_{2} = 15 X 2 - \frac{9.8 X2^2}{2}$

$y_{2} = 10.4m$

<u>Fourth step: </u>we determine the y-component of the center mass of the two balls

$y = \frac{m_{1}y_{1} +m_{2}y_{2}}{m_{1} +m_{2} }$

$y = \frac{(0.5)X(5.4) +(0.25) X (10.4)}{(0.5 +0.25) }$

y = 7.1m

<u>Fifth step:</u> we solve B part of the question; velocity of the center mass of the two balls

$Velocity = \frac{distance of center mass of the two balls (y)}{time}$

$velocity = \frac{7.1 m}{2 s}$

velocity = 3.55m/s

<u>Sixth step:</u> we solve C part of the question; acceleration of the center mass of the two balls

$acceleration = \frac{velocity}{time}$

$acceleration = \frac{3.55}{2}$

acceleration = 1.775 m/s^2

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3 years ago