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Digiron [165]
3 years ago
9

A quantity of an ideal gas is kept in a rigid container of constant volume. If the gas is originally at a temperature of 28 °C,

at what temperature (in °C) will the pressure of the gas triple from its base value?
Physics
1 answer:
Gnoma [55]3 years ago
8 0

Answer:

T_2=630^{\circ}C'

Explanation:

Original temperature of the gas, T_1=28^{\circ}C=301\ K

From the ideal gas equation,

P_1V_1=nRT_1

Since,

P_2=3P_1

nRT_2=3(nRT_1)

T_2=3T_1

T_2=3\times 301

T_2=903\ K

or

T_2=630^{\circ}C

So, the new temperature of the gas is 630 degree Celsius. Hence, this is the required solution.

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Well if the ship was in space their shouldn’t be a loud bang. Because you can’t hear anything in space
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What are some types of landforms on Earth’s surface?<br><br><br><br> PLS ANSWER QUICK 11 POINTS
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plateau, mountains, hills, plains

5 0
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A fish inside the water 12cm below the surface looking up through the water sees the outside world contained in a circular horiz
serg [7]

Answer:

13.6 cm

Explanation:

From Snell's law:

n₁ sin θ₁ = n₂ sin θ₂

In the air, n₁ = 1, and light from the horizon forms a 90° angle with the vertical, so sin θ₁ = sin 90° = 1.

Given n₂ = 4/3:

1 = 4/3 sin θ

sin θ = 3/4

If x is the radius of the circle, then sin θ is:

sin θ = x / √(x² + 12²)

sin θ = x / √(x² + 144)

Substituting:

3/4 = x / √(x² + 144)

9/16 = x² / (x² + 144)

9/16 x² + 81 = x²

81 = 7/16 x²

x ≈ 13.6

4 0
4 years ago
Which one of the following statements concerning the momentum of a system when the net force acting on the system is equal to ze
Alex

Answer:

D

Explanation:

According to newton's 2nd law rate of change of momentum is directly proportional to the force applied on the body. Since, net Force is zero this means momentum did not change or momentum of the body remained constant.

Hence, the system have constant value of momentum. Therefore, option D is correct.

5 0
3 years ago
SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the
Firlakuza [10]

Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

H = 3 m

Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)

Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

5 0
3 years ago
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