How do you describe sound? (SELECT ALL THAT APPLY.) PLEZ HELP MEH!!

Consider the expansion of a gas at a constant temperature in a water-cooled piston-cylinder system. The constant temperature is

Leona [35]

Answer:

$Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})$

Explanation:

According to the first thermodynamic law, the energy must be conserved so:

$dQ = dU - dW$

Where Q is the heat transmitted to the system, U is the internal energy and W is the work done by the system.

This equation can be solved by integration between an initial and a final state:

(1) $\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU - \int\limits^1_2 {} \, dW$

As per work definition:

$dW = F*dr$

For pressure the force F equials the pressure multiplied by the area of the piston, and considering dx as the displacement:

$dW = PA*dx$

Here A*dx equals the differential volume of the piston, and considering that any increment in volume is a work done by the system, the sign is negative, so:

$dW = - P*dV$

So the third integral in equation (1) is:

$\int\limits^1_2 {- P} \, dV$

Considering the gas as ideal, the pressure can be calculated as $P = \frac{n*R*T}{V}$ , so:

$\int\limits^1_2 {- P} \, dV = \int\limits^1_2 {- \frac{n*R*T}{V}} \, dV$

In this particular case as the systems is closed and the temperature constant, n, R and T are constants:

$\int\limits^1_2 {- \frac{n*R*T}{V}} \, dV = -nRT \int\limits^1_2 {\frac{1}{V}} \, dV$

Replacion this and solving equation (1) between state 1 and 2:

$\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU + nRT \int\limits^1_2 {\frac{1}{V}} \, dV$

$Q_{2} - Q_{1} = U_{2} - U_{1} + nRT(ln V_{2} - ln V_{1})$

$Q_{2} - Q_{1} = U_{2} - U_{1} + nRT ln \frac{V_{2}}{V_{1}}$

The internal energy depends only on the temperature of the gas, so there is no internal energy change $U_{2} - U_{1} = 0$ , so the heat exchanged to the system equals the work done by the system:

$Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})$

4 0

3 years ago

Static electricity is the result of

grin007 [14]

Answer:

Explanation:

Static electricity is the result of an imbalance between negative and positive charges in an object. These charges can build up on the surface of an object until they find a way to be released or discharged.

8 0

2 years ago

In Millikan's oil drop experiment, if the electricfield between the plates was of just the right magnitude, it wouldexactly bala

Vilka [71]

Answer:

The electric field to balance the weight is approximately equals to 3.49x10^5 Newton/Coulumb

Explanation:

In order to be stationary position, magnitude of the total force due to electric field should be equal to the gravitational force that is, $|F_{electrostatic}| = |F_{gravitational}|$

where

$F_{electrostatic}=q.E$

where q is the charge of the droplet and E is the electric field. On the other hand

$F_{gravitational}=m.g=V.d.g=\frac{4}{3}.\pi.r^3.d.g$

where m,V ,d and r are the mass, volume, density and radius of the oil droplet respectively and g is the gravitational acceleration (g=9,80665 m/sn^2). By using the first equation and solving it for the electric field we can write,

$q.E=\frac{4}{3}.\pi.r^3.d.g\\E=\frac{4}{3}.\pi.\frac{r^3.d.g}{q}\\E=\frac{4}{3}.\pi\frac{(1,6x10^{-4})^3.(0,85x10^{-3}).(9,81)}{1,6x10^{-19}}\\E\approx 3,49x10^{5} (N/C)$

(Note: d=0.85 x 10^-3 kg/cm^3 and the unit of electric field is Newton per coulumb (N/C))

8 0

3 years ago

I have to invent something for people 65pluse so I need to know what are something old people would bye.

erastova [34]

Phones would be a great thing.

3 0

2 years ago

Q1 Car B is traveling a distance d ahead of car A. Both

MAVERICK [17]

Answer:

The minimum distance, d, between the cars so as to avoid collision is approximately 5.1304 feet

Explanation:

The given parameters are;

The distance of car B ahead of car A = d

The initial speed of both car A and car B = 80 ft./s

The deceleration of car B due to the application of the breaks = 10 ft./s²

The reaction time of the driver of car A = 0.65 s

The deceleration of the driver of car A = 17 ft./s²

To avoid collision, we have;

v² = u² - 2·a·s

Where;

v = The final velocity

u = The initial velocity

a = The acceleration

s = The distance covered

Where the car B comes to rest, we have, v = 0, therefore;

0² = u² - 2·a·s

u² = 2·a·s

s = u²/(2·a)

The distance travelled by car B before coming to rest, s, is therefore;

s = (80 ft./s)²/(2 × 10 ft./s²) = 320 ft.

The distance travelled by car A, after car B applies brakes, is given as follows;

The distance travelled at constant speed = u × Reaction time

∴ The distance travelled at constant speed = 80 ft./s × 0.65 s = 52 ft.

The distance, s, travelled by car A under deceleration is given as follows;

s = (80 ft./s)²/(2 × 17 ft./s²) = 188.24 ft.

s = 80·(t - 0.65) - (1/2)·17·(t - 0.65)² + 80 × 0.65 = d + 80·t - (1/2)·10·t²

∴ d = 80·(t - 0.65) - (1/2)·17·(t - 0.65)² + 80 × 0.65 - (80·t - (1/2)·10·t²)

d = -3.5·t² + 11.05·t - 3.59125

dd/dt = -7·t + 11.05 = 0

The time for the maximum distance for the cars to collide, t = 11.05/7

The maximum distance between the cars for collision, d = -3.5 × (11.05/7)² + 11.05 × (11.05/7) - 3.59125 ≈ 5.1304

Therefore, for no collision, the distance between the cars, d, should be approximately more than 5.1304 feet

5 0

3 years ago