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viva [34]
3 years ago
9

An expanding gas does 153 J of work on its surroundings at a constant pressure of 1.01 atm. If the gas initially occupied 68.0 m

L, what is the final volume of the gas?
Physics
1 answer:
Natalija [7]3 years ago
3 0

Answer:

69.5mL

Explanation:

We are given that

Work done by gas on surroundings=w=153 J=\frac{153}{101.3}Latm

1Latm=101.3J

Pressure =P=1.01atm

Initial volume of gas=68 mL

We have to find the final volume of the gas.

We know that

Work done on constant pressure

w=P\Delta V=P(V_2-V_1)

Using the formula

\frac{153}{101.3}=1.01(V_2-68)

V_2-68=\frac{153}{101.3\times 1.01}=1.5

V_2=68+1.5=69.5mL

Hence, the final volume of gas=69.5mL

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Why is mercury not suitable as the liquid in the U-tube? (the relationship between density of the liquid and the pressure in the
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Relation between density and pressure

Pressure = force/area

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Pressure = mass * acceleration/ area

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A 12.0-kg shell is launched at an angle of 55.0 ∘ above the horizontal with an initial speed of 150 m/s. when it is at its highe
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How can a calculated height be greater than an actual height?
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2 years ago
Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th
Strike441 [17]

Answer:

E=3.5(8.98*10^{6}x-2.69*10^{15}t)

B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)

Explanation:

The electric field equation of a electromagnetic wave is given by:

E=E_{max}(kx-\omega t) (1)

  • E(max) is the maximun value of E, it means the amplitude of the wave.
  • k is the wave number
  • ω is the angular frequency

We know that the wave length is λ = 700 nm and the peak electric field magnitude of 3.5 V/m, this value is correspond a E(max).

By definition:

k=\frac{2\pi}{\lambda}            

k=8.98*10^{6} [rad/m]      

And the relation between λ and f is:                

c=\lambda f

f=\frac{c}{\lambda}

f=\frac{3*10^{8}}{700*10^{-9}}

f=4.28*10^{14}

The angular frequency equation is:

\omega=2\pi f

\omega=2\pi*4.28*10^{14}

\omega=2.69*10^{15} [rad/s]

Therefore, the E equation, suing (1), will be:

E=3.5(8.98*10^{6}x-2.69*10^{15}t) (2)

For the magnetic field we have the next equation:

B=B_{max}(kx-\omega t) (3)

It is the same as E. Here we just need to find B(max).

We can use this equation:

E_{max}=cB_{max}

B_{max}=\frac{E_{max}}{c}=\frac{3.5}{3*10^{8}}

B_{max}=1.17*10^{-8}T

Putting this in (3), finally we will have:

B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t) (4)

I hope it helps you!

8 0
3 years ago
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