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marishachu [46]
3 years ago
12

A gas is compressed at a constant pressure of 0.800 atm from 12.00 L to 3.00 L. In the process, 390 J of energy leaves the gas b

y heat.
a) What is the work done on the gas?
b) What is the change in its internal energy?
Physics
1 answer:
Andru [333]3 years ago
5 0

Answer:

a)W= - 720 J

b)ΔU= 330 J

Explanation:

Given that

P = 0.8 atm

We know that 1 atm = 100 KPa

P = 80 KPa

V₁ = 12 L = 0.012 m³       ( 1000 L = 1 m³)

V₂ = 3 L = 0.003 m³

Q= - 390 J ( heat is leaving from the system )

We know that work done by gas given as

W = P (V₂ -V₁ )

W= 80 x ( 0.003 - 0.012 ) KJ

W= - 0.72 KJ

W= - 720 J    ( Negative sign indicates work done on the gas)

From first law of thermodynamics

Q = W + ΔU

ΔU=Change in the internal energy

Now by putting the values

- 390 = - 720 + ΔU

ΔU= 720 - 390  J

ΔU= 330 J

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In a physics laboratory experiment, a coil with 200 turns enclosing an area of 13.1 cm2 is rotated during the time interval 3.10
sergij07 [2.7K]

Answer:

A)\Phi=83.84\times 10^{-9}

B)\Phi=0 Wb

C)emf=5.4090\times 10^{-4}V

Explanation:

Given that:

  • no. of turns i the coil, n=200
  • area of the coil, a=13.1 \times 10^{-4}\,m^2
  • time interval of rotation, t=3.1\times 10^{-2}\,s
  • intensity of magnetic field, B=6.4\times 10^{-5}\,T

(A)

Initially the coil area is perpendicular to the magnetic field.

So, magnetic flux is given as:

\Phi=B.a\,cos \theta..................................(1)

\theta is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.

\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 0^{\circ}

\Phi=83.84\times 10^{-9} Wb

(B)

In this case the plane area is parallel to the magnetic field i.e. the area vector is perpendicular to the magnetic field.

∴  \theta=90^{\circ}

From eq. (1)

\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 90^{\circ}

\Phi=0 Wb

(C)

According to the Faraday's Law we have:

emf=n\frac{B.a}{t}

emf=\frac{200\times 6.4\times 10^{-5}\times 13.1 \times 10^{-4}}{3.1\times 10^{-2}}

emf=5.4090\times 10^{-4}V

7 0
3 years ago
The force of gravitation between two spherical bodies is Gm1
malfutka [58]

Answer:

r is the separation between the two spherical bodies

3 0
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The amplitude of a lightly damped oscillator decreases by 3.42% during each cycle. What percentage of the mechanical energy of t
frozen [14]

Answer:

The percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%

Explanation:

Mechanical energy (Potential energy, PE) of the oscillator is calculated as;

PE = ¹/₂KA²

During the first oscillation;

PE₁ = ¹/₂KA₁²

During the second oscillation;

A₂ = A₁ - 0.0342A₁ = 0.9658A₁

PE₂ = ¹/₂KA₂²

PE₂ = ¹/₂K (0.9658A₁)²

PE₂ = (0.9658²)¹/₂KA₁²

PE₂ = (0.9328)¹/₂KA₁²

PE₂ = 0.9328PE₁

Percentage of the mechanical energy of the oscillator lost in each cycle;

Change \ in \ percent= \frac{PE_2 - PE_1}{PE_1} \\\\Change \ in \ percent= \frac{0.9328PE_1 -PE_1}{PE_1} *100\\\\Change \ in \ percent= \frac{-0.0672PE_1}{PE_1}*100 \\\\Change \ in \ percent= -0.0672*100\\\\Change \ in \ percent= -6.72 \ \% \\\\Loss \ in \ percent= 6.72 \ \%

Therefore, the percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%

4 0
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The number of employees for a certain company has been decreasing each year by 4%. If the company currently has 650 employees an
nika2105 [10]
The answer would probably be 649.96
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3 years ago
Pls help me I don’t get it :(
Ksju [112]

Answer:

2nd and 4th

Explanation:

4 0
3 years ago
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