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natka813 [3]
1 year ago
15

In this problem you are to consider an adiabaticexpansion of an ideal diatomic gas, which means that the gas expands with no add

ition or subtraction of heat.Assume that the gas is initially at pressure p0, volume V0, and temperature T0. In addition, assume that the temperature of the gas is such that you can neglect vibrational degrees of freedom. Thus, the ratio of heat capacities is γ=Cp/CV=7/5.Note that, unless explicitly stated, the variable γ should not appear in your answers--if needed use the fact that γ=7/5 for an ideal diatomic gas.Part AFind an analytic expression for p(V), the pressure as a function of volume, during the adiabatic expansion.Express the pressure in terms of V and any or all of the given initial values p0, T0, and V0.Correctp(V) = p0(V0V)75Part BAt the end of the adiabatic expansion, the gas fills a new volume V1, where V1>V0. Find W, the work done by the gas on the container during the expansion.Express the work in terms of p0, V0, and V1. Your answer should not depend on temperature.Part CFind ΔU, the change of internal energy of the gas during the adiabatic expansion from volume V0 to volume V1.Express the change of internal energy in terms of p0, V0, and/or V1.Need Part B and C

Physics
1 answer:
rosijanka [135]1 year ago
5 0

Answer:

A: P=Po(Vo/V)^y

B: W=5/2Po(Vo-Vo^7/5 Vi^-2/5)

C: ∆V= 5/2Po[Vo^7/5 Vi^-2/5 -Vo]

Explanation:

Attached is the solution

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Charge q1 is distance r from a positive point charge Q. Charge q2=q1/3 is distance 2r from Q. What is the ratio U1/U2 of their p
worty [1.4K]

We have that The ratio U1/U2 of their potential energies due to their interactions with Q is

  • U1/U2=6
  • U1/U2=6

From the question we are told that

Question 1

Charge q1 is distance r from a positive point charge Q.

Question 2

Charge q2=q1/3 is distance 2r from Q.

Charge q1 is distance s from the negative plate of a parallel-plate capacitor.

Charge q2=q1/3 is distance 2s from the negative plate.

Generally the equation for the potential energy  is mathematically given as

U=\frac{-k*qQ}{r}

Therefore

The Equations of U1 and U2 is

For U1

U1=\frac{-k*q_1Q}{r}

For U2

U2=\frac{-k*q_1Q}{3*2r}

Since

U is a function of q and  q2=q1/3

Therefore

U1/U2=6

For Question 2

For U1

U1=\frac{-k*q_1Q}{s}\\\\For U2\\\\U2=\frac{-k*q_1Q}{3*2r}

Therefore

U1/U2=6

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7 0
1 year ago
A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,00
Aleksandr-060686 [28]

Answer:

Explanation:

We have the following relation between power, P and intensity, I

I = P/(4*pi*r^2)

= 10^3/(4*pi*(35000*10^3))

= 6.5*10^-14 W/M^2

We also have the following relationship between electric field and electromagnetic radiation thus

I = (ceE^2)/2

Hence E = \sqrt{2I/ce}

substituting the values of I, c and e, we have

7*10^-6 V/m

3 0
1 year ago
the brightest , hottest, and most massive stars are the brilliant blue stars designated as spectral class O. if a class O star w
4vir4ik [10]

The speed is 0.956 m / s.

<u>Explanation</u>:

The kinetic energy is equal to the product of half of an object's mass, and the square of the velocity.

                   K.E = 1/2 \times m \times v^{2}

where K.E represents the kinetic energy,

           m represents the mass,

            v represents the velocity.

                  K.E = 1/2 \times m \times v^{2}

    1.10 \times 10^42 = 1/2 \times 3.26 \times 10^31 \times v^{2}

                     v^{2} = (1.10 \times 10^42 \times 2) / (3.26 \times 10^31)

                     v = 0.956 m / s.

6 0
2 years ago
. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it
kap26 [50]

Answer:

29.4 N/m

0.1  

Explanation:

a) From the restoring Force we know that :  

F_r = —k*x  

the gravitational force :  

F_g=mg  

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force  

xi , x2 are the displacement made by the two masses.

Givens:

<em>m1 = 1.29 kg</em>

<em>m2 = 0.3 kg  </em>

<em>x1   = -0.75 m  </em>

<em>x2 = -0.2 m </em>

<em>g   = 9.8 m/s^2  </em>

Plugging known information to get :

F_r =F_g

-k*x1 + k*x2=m1*g-m2*g

k=29.4 N/m

b) To get the unloaded length 1:  

l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m  

Plugging known infromation to get :

l= x1 — (F_1/k)  

= 0.1  

 

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VladimirAG [237]

Answer:

lily's speed would be twice john's speed

7 0
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