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lilavasa [31]
3 years ago
13

What’s the correct answer, d is false .

Physics
1 answer:
Alina [70]3 years ago
4 0

I'd Say B :) Your Welcome LOL!!!!!

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A mole of a monatomic ideal gas at point 1 (101 kPa, 5 L) is expanded adiabatically until the volume is doubled at point 2. Then
Paha777 [63]

Answer:

(a). Check attachment.

(b). 280.305 J.

(c). 31.81 kpa; 38.26K.

(d). 24.05K.

(e). 24.05k; 40kpa.

(f). -138.6J.

Explanation:

(a). Kindly check the attached picture for the diagram showing the four process.

1 - 2 = adiabatic expansion process.

2 - 3 = Isochoric process.

3 - 4 = isothermal process.

4 - 1 = isochoric process.

(b). Recall that the process from 1 to is an adiabatic expansion process.

NB: b = 5/3 for a monoatomic gas.

Then, the workdone = (1/ 1 - 1.66) [ (p1 × v1^b)/ v2^b × v2 - (p1 × v1)].

= ( 1/ 1 - 5/3) [ (101 × 5^5/3) × 10^1 -5/3] - 101 × 5.

Thus, the workdone = 280.305 J.

(c). P2 = P1 × V1^b/ V2^b = 101 × 5^5/3/ 10^5/3 = 31.81 kpa.

T2 = P2 × V2/ R × 1 = 31.81 × 10/ 8.324 = 38.36k.

(d). The process 2 - 3 is an Isochoric process, then;

T3 = T2/P2 × P3 = 38.26/ 31.82 × 20 = 24.05K.

(e). The process 3 - 4 Is an isothermal process. Then, the temperature at 4 will be the same temperature at 3. Tus, we have the temperature; point 3 = point 4 = 24.05k.

The pressure can be determine as below;

P4 = P3 × V3/ V4 = 20 × 10/ 5 = 200/ 5 = 40 kpa.

(f) workdone = xRT ln( v4/v3) = 1 × 8.314 × 24.05 × ln (5/10) = - 138.6 J

6 0
3 years ago
What are the characteristics of a nebulae? (Select all that apply.)
erica [24]

Answer:

B. contain hydrogen

C. clouds of gas and dust

E. needed to create a star

Explanation:

A star is a giant astronomical or celestial object that is comprised of a luminous sphere of plasma, binded together by its own gravitational force.

Some of the examples of stars are; Vega, Sun (closest to planet Earth), Antares, Betelgeus, Canopus, etc.

Stars are typically made up of two (2) main hot gas, Hydrogen (H) and Helium (He). The chronological order in which the formation of a star occur are;

1. Gravity pulls gas and dust together to form dense cores.

2. A protostar forms as mass increases.

3. Nuclear fusion begins under high pressure.

Scientists have been able to understand and discover that, gravity pulled materials (low-density cloud of interstellar gas and dust known as a nebula) together forming the planetary bodies in our solar system.

A dark nebula can be defined as an interstellar cloud that is so dense as a result of high concentration of gas and dust and as such it obscures the visible wavelengths of light from stars behind it, thus appearing completely opaque (dark patch) in front of a bright emission nebula or in regions having plenty stars.

The characteristics of a nebulae are;

I. It contain hydrogen.

II. Clouds of gas and dust

III. It is needed to create a star.

7 0
3 years ago
The potential difference across a variable resistor is 11V and the current flowing through it is 0.4A.
klemol [59]
The resistance is 27.5 ohms
5 0
3 years ago
Read 2 more answers
Objects in space that are moving at a constant velocity in a straight line ___________.
nikdorinn [45]
The best answer is A) <span>keep moving at a constant velocity until some forces act on them

As the man you're probably tired of hearing about said:

"Every object persists in its state of rest or in uniform motion in a straight line unless a new force acts upon it" 
This is Isaac Newton's 1st law of motion, or the law of inertia. 

Put more simply, objects in motion tend to stay in motion, and tend the maintain the same velocity (direction and speed) and objects at rest tend to stay at rest. </span>
6 0
2 years ago
Read 2 more answers
Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on
Natasha2012 [34]

Answer:

8.8\times 10^{-6} W/m^2

Explanation:

We are given that

Wavelength,\lambda=571 nm=571\times 10^{-9} m

1 nm=10^{-9} m

R=75 cm=\frac{75}{100}=0.75 m

1 m=100 cm

d=0.640 mm=0.64\times 10^{-3} m

1 mm=10^{-3} m

a=0.434 mm=0.434\times 10^{-3} m

y=0.830 mm=0.830\times 10^{-3} m

I_0=5.00\times 10^{-4}W/m^2

tan\theta=\frac{y}{R}

\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}

\theta=1.1\times 10^{-3}rad

tan\theta\approx \theta

Because \theta is small.

\phi=\frac{2\pi dsin\theta}{\lambda}

\sin\theta\approx \theta,

Therefore

\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}

\phi=7.74 rad

\beta=\frac{2\pi a\theta}{\lambda}

\beta=5.3 rad

I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2

I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2

I=8.8\times 10^{-6} W/m^2

6 0
3 years ago
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