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stellarik [79]
3 years ago
9

Some hydrogen gas is enclosed within a chamber being held at 200∘c with a volume of 0.0250 m3. the chamber is fitted with a mova

ble piston. initially, the pressure in the gas is 1.50×106pa (14.8 atm). the piston is slowly extracted until the pressure in the gas falls to 0.950×106pa. what is the final volume v2 of the container? assume that no gas escapes and that the temperature remains at 200∘c. enter your answer numerically in cubic meters.
Physics
1 answer:
Mrac [35]3 years ago
4 0

Answer: The final volume V₂ of the container is  0.039 m³.

Explanation:

Since the temperature is constant, the gas would expand isothermally.

For isothermal expansion,

P₁V₁=P₂V₂

Where, P₁ and P₂ are the initial and final pressure and V₁ and V₂ are initial and final volume.

It is given that:

V₁ = 0.0250 m³

P₁ = 1.5 × 10⁶ Pa

P₂ = 0.950 × 10⁶ Pa

V₂ = ?

⇒ 1.5 × 10⁶ Pa × 0.0250 m³ = 0.950 × 10⁶ Pa × V₂

⇒V₂ = 0.039 m³

Hence, the final volume V₂ of the container is  0.039 m³.

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Answer:

the correct answer is C,   281 lb

Explanation:

This is an exercise in fluid mechanics specifically on Pascal's principle, which states that the pressure in a fluid is equal to all points that are at the same depth

             P = \frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }

If we use the subscript 1 for the piston diameter mentor (d = 1.00vm) which is r = 0.05 cm, the force is F1 = 45 lb

          F₂ = \frac{F_{1} }{A_{1} }  F₁

in area of ​​a circules

         A = π r²

we substitute

          F₂ = \frac{r_{2}^2 }{r_{1}^2}  F_{1}

let's calculate

           F2 = 2.50² / 1.00²  45

           F2 =281  lb

therefore the correct answer is C

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Suppose two waves collide, and the temporary combined wave that results is smaller than the original waves. what term best descr
BlackZzzverrR [31]
The collision of the two waves would bring force and that would make the new wave smaller
3 0
3 years ago
A cylinder with initial volume V contains a sample of gas at pressure p. On one end of the cylinder, a piston is let free to mov
Elina [12.6K]

Answer:

The work done on the gas = 2PV

Explanation:

In this question, we need to apply the basic gas laws to determine the word done.

For this question, we need to draw a PV diagram (a pressure-volume diagram), which I have made and attached in the attachment. So please refer to that attachment. I will be using this diagram to solve for the work done on the gas.

So, Please refer to the attachment number 1. where x -axis is of volume and y-axis is of pressure.

As we know that, the work done on the gas is equal to the area under the curve.

W = Area of the triangle

W = 0.5 x ( base) x ( height)

W = 0.5 x (BC) x (AC)

W = 0.5 x (3V-V) x (3P-P)

W = 2PV

Hence, the work done on the gas = 2PV

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3 years ago
In the sum A→+B→=C→, vector A→ has a magnitude of 13.6 m and is angled 40.2° counterclockwise from the +x direction, and vector
il63 [147K]

Answer:

|B|=27.00425726m

\alpha =210.3781372°

Explanation:

Let's use the component method of vector addition:

A_x=13.6cos(40.2)=10.38762599\\A_y=13.6sin(40.2)=8.778224553\\Cx=13.8cos(20.7+180)=-12.90912763\\Cy=13.8sin(20.7+180)=-4.877952844

Now, we know:

C_x=A_x+B_x\\\\C_y=A_y+b_y

So:

B_x=C_x-A_x=-23.29675362\\B_y=C_y-A_y=-13.6561774

Now lets calculate the magnitude of the vector B:

|B|=\sqrt{(B_x)^{2} +(B_y)^{2}  }=27.00425726m

Finally its angle is given by:

\alpha =(arctan(\frac{B_y}{B_x}))+180=30.37813438+180=210.3781344°

Keep in mind that I added 180 to the angles of C and B to find the real angles measured from the + x axis counter-clock wise.

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Newton's third law: For every action, there is an opposite and equal reaction.

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