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Allisa [31]
3 years ago
9

A vertical cylinder with a heavy piston contains air at 300 K. The initial pressure is 2.0 x 105 Pa and the initial volume is 0.

35 m3 . Take the molar mass of air as 28.9 g/mol and assume Cv= 5 2R. (A) Find the specific heat of air at constant volume in units of J kg·K . (5 pts) (B) Calculate the mass of air in the cylinder. (5 pts) (C) Suppose the piston is fixed. Find the energy input required to raise the temperature to 700 K. (5 pts). (D) Assume again the conditions of the initial state and assume the piston is free to move. Find the energy input to raise the temperature to 700 K. (10 pts)

Physics
1 answer:
aleksklad [387]3 years ago
8 0

Find answers and explanations in the attachments

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A plastic bottle partially filled with water floats on water, even though the density of the plastic (1.2 g/cc) is more than tha
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Answer:

True the plastic will float because of the principle of flotation or buoyancy

Explanation:

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Buoyancy  is the upward force/upthrust experienced by a body immersed totally or partially in a liquid.

According to the principle of flotation:

<em>"when a body is totally or partially immersed in liquid it experiences an upthrust which is equal to the volume of fluid displaced"</em>

The plastic will float due to the fact the average density of the total volume of the plastic and the air inside it is less than the same volume of water it is floating in

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A 5.09 × 1014-hertz electromagnetic wave is traveling through a transparent medium. The main factor that determines the speed of
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6 0
3 years ago
A car starts from rest and accelerates uniformly at 2.0 m/s2 toward the north. A second car starts from rest 4.0 s later at the
yKpoI14uk [10]

Answer:

the correct solution is 13 s

Explanation:

This is a kinematic problem, let's use accelerated rectilinear motion relationships.

For the first car it has an accelerometer of 2.0 m/s²

       x = v₀₁ t + ½ a₁ t²

The second car leaves the same point, but 4.0 seconds later

       x = v₀₂ (t-4) + ½ a₂ (t-4)²

With this form we use the same time for both cars.

The initial speeds are zero for both vehicles leave the rest, at the point where they are located has the same position

        x = ½ a₁ t²

        x = ½ a₂ (t-4)²

Let's solve

       a₁  t² = a₂ (t-4)²

      a₁/a₂ t² = t² -2 4 t + 16

      t² (1- 2.0 / 4.0) - 8 t +16

      t² 0.5 - 8 t +16 = 0

      t² -16 t + 32 = 0

Let's solve the second degree equation

     t = [16 ±√( 16² - 4 32)] / 2  

     t = ½ (16 ± 11,3)

Solutions

     t1 = 13.66 s

     t2 = 2.34 s

These are the mathematical solutions for the meeting point, but car 2 leaves after 4 seconds, so the only solution is 13.66 s

the correct solution is 13 s, if you have to select one the nearest 12s

6 0
3 years ago
Using the equation zeff=z−s and assuming that core electrons contribute 1.00 and valence electrons contribute 0.00 to the screen
DIA [1.3K]
Thank you for posting your question here at brainly. Below is the solution. I hope the answer will help. 

<span>Cl^- 1s^2 2s^2p^6 3s^2 3p^6 1s^2 2s^2p^6 S = 10; 3s^2 3p^6 S = 0 </span>
<span>Zeff = Z-S = 17- 10 =7 </span>
<span>K^+ 1s^2 2s^2p^6 3s^2 3p^6; 1s^2 2s^2p^6 S = 10; 3s^2 3p^6 S = 0 </span>
<span>Zeff = Z-S = 19- 10 = 9 
</span>
S = 2 + 6.8 + 2.45 = 11.25 
<span>Zeff(Cl^-) = 17 – 11.25 = 5.75 </span>
<span>K^+ 1s^2 2s^2p^6 3s^2 3p^6 same S as for Cl^- but Z increases by 2 hence </span>
<span>Zeff(K^+) = 19 - 11.25 = 7.75</span>
5 0
3 years ago
Read 2 more answers
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