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Ahat [919]
4 years ago
8

In which process will the internal energy of the system NOT change? a. An adiabatic expansion of an ideal gas.

Physics
1 answer:
marusya05 [52]4 years ago
3 0

Answer:

B. An isothermal compression of an ideal gas.

Explanation:

The internal energy of an ideal gas is just function of the temperature; it does not matter what other thermodynamic property changes, if the temperature does not change, the internal energy neither does. That is just for ideal gases; real gases behaviour is not like that. All of the other options bring with them an increase or decrease of the temperature:

For A, the temperature will decrease because the gas will do work as it expands, converting part of his internal energy to work.

For C, the temperature will increase, because given PV=nRT, if the volume increases (expansion) and the pressure is constant, the temperature must increase to satisfy the equation.

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What is Newton’s second law of motion
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"<em>F = dP/dt. </em> The net force acting on an object is equal to the rate at which its momentum changes."

These days, we break up "the rate at which momentum changes" into its units, and then re-combine them in a slightly different way.  So the way WE express and use the 2nd law of motion is

"<em>F = m·A.</em>  The net force on an object is equal to the product of the object's mass and its acceleration."

The two statements say exactly the same thing. You can take either one and work out the other one from it, just by working with the units.

8 0
3 years ago
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The air strips at high altitude airports are longer than those at sea level. What would be the reason for this? Explain using sc
Eddi Din [679]

Answer:

The lower air density at high altitudes means that a faster air speed is need for a given amount of lift. In order to get the higher air speed, an airplane must accelerate for a longer amount of time, and thus a longer runway is required.

Explanation:

6 0
3 years ago
If a car can go from 0.0 to 60.0 mi/hr in 8.0 seconds, what would be its final speed after 5.0 seconds if its
o-na [289]
a =  \frac{60 }{8}  = 7,5 \ mph \ gained \ each\ second \\ final \ speed: 97.5 \ mph \ [ 50 + 7,5 * 5]
4 0
3 years ago
A mass of 4 kg is traveling over a quarter circular ramp with a radius of 10 meters. At the bottom of the incline the mass is mo
amid [387]

Answer:

499.7 J

Explanation:

Since total mechanical energy is conserved,

U₁ + K₁ + W₁ = U₂ + K₂ + W₂ where U₁ = potential energy at bottom of incline = mgh₁, K₁ = kinetic energy at bottom of incline = 1/2mv₁² and W₁ = work done by friction at bottom of incline, and U₂ = potential energy at top of incline = mgh₂, K₁ = kinetic energy at top of incline = 1/2mv₂² and W₂ = work done by friction at top of incline. m = mass = 4 kg, h₁ = 0 m, v₁ = 21.3 m/s, W₁ = 0 J, h₂ = radius of circular ramp = 10 m, v₂ = 2.8 m/s, W₂ = unknown.

So, U₁ + K₁ + W₁ = U₂ + K₂ + W₂

mgh₁ + 1/2mv₁²  + W₁ = mgh₂ + 1/2mv₂²  + W₂

Substituting the values of the variables into the equation, we have

mgh₁ + 1/2mv₁²  + W₁ = mgh₂ + 1/2mv₂²  + W₂

4 kg × 9.8 m/s²(0) + 1/2 × 4 kg × (21.3 m/s)²  + 0 = 4 kg × 9.8 m/s² × 10 m + 1/2 × 4 kg × (2.8 m/s)²  + W₂

0 + 2 kg × 453.69 m²/s² = 392 kgm²/s² + 2 kg × 7.84 m²/s²  + W₂

907.38 kgm²/s² = 392 kgm²/s² + 15.68 kgm²/s²  + W₂

907.38 kgm²/s² = 407.68 kgm²/s² + W₂

W₂ = 907.38 kgm²/s² - 407.68 kgm²/s²

W₂ = 499.7 kgm²/s²

W₂ = 499.7 J

Since friction is a non-conservative force, the work done by all the non-conservative forces is thus W₂ = 499.7 J

4 0
3 years ago
Should the size of the test charge compare to the amount of charge that produces the field
Alla [95]

Answer:

20000000000)0000000000000

4 0
3 years ago
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