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guajiro [1.7K]
3 years ago
6

A cylinder with a piston contains 0.200 mol of nitrogen at 1.50×105 Pa and 320 K . The nitrogen may be treated as an ideal gas.

The gas is first compressed isobarically to half its original volume. It then expands adiabatically back to its original volume, and finally it is heated isochorically to its original pressure. Find the heat added to the gas during the final heating. Find the internal-energy change of the gas during the final heating.
Physics
1 answer:
Alja [10]3 years ago
3 0

Answer:

Q = -105 J

Also we know that for cyclic process change in internal energy is always ZERO

Explanation:

First gas is compressed isobarically such that its volume is half of initial volume

So its temperature is also half

So heat given in this process is given as

Q = nC_p \Delta T

for diatomic gas we have

C_p = \frac{7}{2} R

so we will have

Q = 0.200(\frac{7}{2}R)(160 - 320)

Q = -930.7 J

Now in adiabatic process heat is not transferred

so in this process

Q = 0

so we have

T_1V_1^{1.4-1} = T_2V_2^{1.4-1}

(160)(\frac{V}{2})^{0.4} = T_2(V)^{0.4}

T_2 = 121.26 K

Now it is again reached to original pressure

so temperature will become initial temperature

so heat given in that part

Q_3 = nC_v\Delta T

here we know that

C_v = \frac{5}{2}R

Q_3 = (0.200)(\frac{5}{2}R)(320 - 121.26)

Q_3 = 825.76 J

So total heat given to the system is

Q = -930.7 + 0 + 825.76

Q = -105 J

Also we know that for cyclic process change in internal energy is always ZERO

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Norma-Jean [14]

Answer:

  3×10^-12 C  

Explanation:

The total of the three charges is ...

   (-3 +8 +4)×10^-12 C = 9×10^-12 C

Assuming the charge is equally distributed between the balls when they are brought in contact, the charge on each ball will be ...

  (9/3)×10^-12 C = 3×10^-12 C

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4. An object is moving with an initial velocity of 9 m/s. It accelerates at a rate of 1.5
Mazyrski [523]

Answer:

11.87 ms⁻¹

Explanation:

You can use the kinematic equation

v² = u² + 2as

Where v = final velocity

          u = initial velocity

          a = acceleration

          s = displacement

v² = 9² + 2×1.5×20

So you get v = 11.87 ms⁻¹

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• Describe the importance of a commonly used measuring system. Be prepared to give two examples of why having a common system of
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3 years ago
The James Webb Space Telescope is positioned around 1.5 million kilometres from the Earth on the side facing away from the Sun.
Bad White [126]

The angular velocity depends on the length of the orbit and the orbital

speed of the telescope.

Response:

First question:

  • The angular velocity of the telescope is approximately <u>0.199 rad/s</u>

Second question:

  • The telescope should accelerates away by approximately F = <u>0.0005·m </u>

Third question:

  • <u>The pulling force between the Earth and the satellite</u>

<h3>What equations can be used to calculate the velocity and forces acting on the telescope?</h3>

The distance of the James Webb telescope from the Sun = 1.5 million kilometers from Earth on the side facing away from the Sun

The orbital velocity of the telescope = The Earth's orbital velocity

First question:

Angular \ velocity = \mathbf{\dfrac{Angle \ turned}{Time \ taken}}

The orbital velocity of the Earth = 29.8 km/s

The distance between the Earth and the Sun = 148.27 million km

The radius of the orbit of the telescope = 148.27 + 1.5 = 149.77

Radius of the orbit, r = 149.77 million kilometer from the Sun

The length of the orbit of the James Webb telescope = 2 × π × r

Which gives;

r = 2 × π × 149.77 million kilometers ≈ 941.03 million kilometers

Therefore;

Angular \ velocity = \dfrac{29.8}{941.03}\times 2 \times \pi \approx 0.199

  • The angular velocity of the telescope, ω ≈ <u>0.199 rad/s</u>

Second question:

Centrifugal force force, F_{\omega} = m·ω²·r

Which gives;

F_{\omega} = m \cdot \dfrac{28,500^2 \, m^2/s^2}{149.77 \times 10^9 \, m} \approx 0.0054233 \cdot m

Gravitational \ force,  F_G = \mathbf{G \cdot \dfrac{m_{1} \cdot m_{2}}{r^{2}}}

Universal gravitational constant, G = 6.67408 × 10⁻¹¹ m³·kg⁻¹·s⁻²

Mass of the Sun = 1.989 × 10³⁰ kg

Which gives;

F_G = 6.67408 \times 10^{-11} \times \dfrac{1.989 \times 10^{30} \times m}{149.77 \times 10^9} \approx   0.00592 \cdot m

Which gives;

F_{\omega} < F_G, therefore, the James Webb telescope has to accelerate away from the Sun

F = \mathbf{F_{\omega}} - \mathbf{F_G}

The amount by which the telescope accelerates away is approximately 0.00592·m - 0.0054233·m ≈ <u>0.0005·m (away from the Sun)</u>

Third part:

Other forces include;

  • <u>The force of attraction between the Earth and the telescope </u>which can contribute to the the telescope having a stable orbit at the given speed.

Learn more about orbital motion here:

brainly.com/question/11069817

3 0
3 years ago
A mass of 790 kg is hanging from a crane (neglect the mass of the cable and the hook). While the mass is being lowered, it is sp
Margarita [4]

Answer:

Explanation:

Given

mass of object hanging from crane is m=790\ kg

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T=790\times 7.1

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