Ask question

Ask question

All categories

erma4kov [3.2K]

3 years ago

11

A gas has a volume of 6 L, a temperature of 70 degrees C, and a pressure of 2 atm. If the gas is compressed to a volume of 4 L,

and the pressure increases to 5 atm, what will be the new temperature of the gas?

1 answer:

Lelu [443]3 years ago

4 0

I would say check the back of the book

You might be interested in

Which type of radio broadcast has a greater broadcast range am or fm

Finger [1]

The answer is FM...That's why all the good stations that people listen to are on FM

3 0

3 years ago

Stephanie lives in California,and her cousin Thomas lives in France. which coordinate system would be most useful if the two wan

qwelly [4]

Answer:

Equatorial

Explanation:

Your welcomee

8 0

3 years ago

Mercury, Mars, and Earth’s moon all show evidence of heavy cratering from meteor impacts while Earth shows much less evidence of

algol [13]

Answer:

D)

Explanation:

4 0

3 years ago

What kind of error would result if you read the liquid volume where the liquid touches the wall of the cylinder rather than at t

shutvik [7]

You would get a wrong calculaton which if you are doing an experiment it can mess with the results

7 0

3 years ago

The linear charge density on the inner conductor is and the linear charge density on the outer conductor is

Lubov Fominskaja [6]

Complete Question

The complete question is shown on the first uploaded image (reference for Photobucket )

Answer:

The electric field is $E = -1.3 *10^{-4} \ N/C$

Explanation:

From the question we are told that

The linear charge density on the inner conductor is $\lambda _i = -26.8 nC/m = -26.8 *10^{-9} C/m$

The linear charge density on the outer conductor is

$\lambda_o = -60.0 nC/m = -60.0 *10^{-9} \ C/m$

The position of interest is r = 37.3 mm =0.0373 m

Now this position we are considering is within the outer conductor so the electric field at this point is due to the inner conductor (This is because the charges on the conductor a taken to be on the surface of the conductor according to Gauss Law )

Generally according to Gauss Law

$E (2 \pi r l) = \frac{ \lambda_i }{\epsilon_o}$

=> $E = \frac{\lambda _i }{2 \pi * \epsilon_o * r}$

substituting values

$E = \frac{ -26 *10^{-9} }{2 * 3.142 * 8.85 *10^{-12} * 0.0373}$

$E = -1.3 *10^{-4} \ N/C$

The negative sign tell us that the direction of the electric field is radially inwards

=> $|E| = 1.3 *10^{-4} \ N/C$

5 0

3 years ago