How many values must a 1-D motion problem give you in order to solve it?

According to the Joule-Thomson experiment, it states that when a gas is expanded adiabatically from higher pressure region to lower pressure region, the change in temperature with respect to change in pressure at constant enthalpy is known as Joule-Thomson coefficient.

The formula will be:

$\mu_{J,T}=(\frac{dT}{dP})_H$

or,

$\mu_{J,T}=(\frac{dT}{dP})_H\approx \frac{\Delta T}{\Delta P}$

As per question the formula will be:

$\mu_{J,T}=\frac{T_2-T_1}{P_2-P_1}$ .........(1)

where,

$\mu_{J,T}$ = Joule-Thomson coefficient of the gas = $0.13K/atm$

$T_1$ = initial temperature = $19.0^oC=273+19.0=292.0K$

$T_2$ = final temperature = ?

$P_1$ = initial pressure = 200.0 atm

$P_2$ = final pressure = 0.95 atm

Now put all the given values in the above equation 1, we get:

$0.13K/atm=\frac{T_2-292.0K}{(0.95-200.0)atm}$

$T_2=266.12K$

Therefore, the final temperature of gas is 266.12 K

5 0

3 years ago

A truck with a mass of 1370 kg and moving with a speed of 12.0 m/s rear-ends a 593 kg car stopped at an intersection. The collis

Elza [17]

Answer:

speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

Explanation:

Given:

mass of truck M = 1370 kg

speed of truck = 12.0 m/s

mass of car m = 593 kg

collision is elastic therefore,

Applying law of momentum conservation we have

momentum before collision = momentum after collision

1370×12 + 0( initially car is at rest) = 1370×v1+ 593×v2 ....(i)

Also for a collision to be elastic,

velocity of approach = velocity of separation

12 -0 = v2-v1 ....(ii)

using (i) and (ii) we have

So speed of car after collision, v2 =16.1 m/s and of the truck, v1 = 4.6 m/s

4 0

3 years ago

Consider the image above. Vi = the initial velocity and Vf = the final velocity. Is there acceleration? Explain your answer.

lutik1710 [3]

Velocity is a vector, and the initial and final ones are in opposite directions.
There must have been acceleration in order to change the direction of motion.

A) No. The initial and final velocities are the same.
This is all wrong, and not the correct choice.
It's "Yes", and the initial and final velocities are NOT the same.

B) Yes. The ball had to slow down in order to change direction.
This is poor, and not the correct choice.
The "Yes" is correct, but the explanation is bad.
Acceleration does NOT require any change in speed.

C) No. Acceleration is the change in velocity. The ball's velocity is constant.
This is all wrong, and not the correct choice.
It's "Yes", there IS acceleration, and the ball's velocity is NOT constant.

D) Yes. Even though the initial and final velocities are the same, there is a change in direction for the ball.
This choice is misleading too.
The "Yes" is correct ... there IS acceleration.
The change in direction is the reason.
The initial and final velocities are NOT the same. Only the speeds are.

3 0

3 years ago

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