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RoseWind [281]
3 years ago
10

Two cylinders A&B at the same temperature contains the same quantity of the same kind of gas. Cylinder A has three times the

volume of cylinder B. What can you conclude about the pressure the gases exert
Physics
1 answer:
GREYUIT [131]3 years ago
3 0

Answer:

pressure in cylinder A must be one third of pressure in cylinder B

Explanation:

We are told that the temperature and quantity of the gases in the 2 cylinders are same.

Thus, number of moles and temperature will be the same for both cylinders.

To this effect we will use the formula for ideal gas equation which is;

PV = nRT

Where;

P is prrssure

V is volume

n is number of moles

T is temperature

R is gas constant

We are told that Cylinder A has three times the volume of cylinder .

Thus;

V_a = 3V_b

For cylinder A;

Pressure = P_a

Volume = 3V_b

Number of moles = n

Thus;

P_a × 3V_b = nRT

For cylinder B;

Pressure = P_b

Volume = V_b

Number of moles = n

Thus,

P_b × V_b = nRT

Combining the equations for both cylinders, we have;

P_a × 3V_b = P_b × V_b

V_b will cancel out to give;

3P_a = P_b

Divide both sides by 3 to get;

P_a = ⅓P_b

Thus, pressure in cylinder A must be one third of pressure in cylinder B

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Answer:

a=2.5\ m/s^2

Explanation:

<u>Motion With Constant Acceleration </u>

It's a type of motion in which the velocity of an object changes uniformly over time.

The equation that describes the change of velocities is:

v_f=v_o+at

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t    = time

Solving the equation for a:

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The ball starts at rest (vo=0) and rolls down an inclined plane that makes it reach a speed of vf=7.5 m/s in t=3 seconds.

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7 0
3 years ago
Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge of
wariber [46]

Answer:

For left = 0  N/C

For right = 0  N/C

At middle = -7.6836 * 10^{-11} \vec{i}  N/C

Explanation:

Given data :-

б =6.8 * 10^{-22} C/ m²

Considering the two thin metal plates to be non conducting sheets of charges.

Electric field is given by

E = \frac{\sigma }{2\varepsilon }

1) To the left of the plate

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(\vec{i})   = 0 N/C.

2) To the right of them.

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(\vec{i})   = 0 N/C.

3) Between them.

\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+  (\frac{\sigma }{2\varepsilon })(-\vec{i}) = (\frac{\sigma }{\varepsilon })(-\vec{i}) = -\frac{6.8 * 10^{-22} }{8.85 * 10 ^{-12} }  \vec{i} =   -7.6836 * 10^{-11} \vec{i} N/C

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Answer:

-10.8m/s^2

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27m/s/2.5=10.8m/s^2

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