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liberstina [14]
2 years ago
5

For the following statements, choose the word or words inside the parentheses that serve to make a correct statement. Each state

ment has at least one and may have more than one correct answer. a. For a sample of an ideal gas, the product pV remains constant as long as the (temperature, pressure, volume, internal energy) is held constant. b. The internal energy of an ideal gas is a function of only the (volume, pressure, temperature). c. The Second Law of Thermodynamics states that the entropy of an isolated system always (increases, remains constant, decreases) during a spontaneous process. d. When a sample of liquid is converted reversibly to its vapor at its normal boiling point, ( q, w, p, V, T, U, H, S, G, none of these) is equal to zero for the system. e. If the liquid is permitted to vaporize isothermally and completely into a previously evacuated chamber that is just large enough to hold the vapor at 1 bar pressure, then ( q, w, U, H, S, G ) will be smaller in magnitude than for the reversible vaporizatio
Physics
1 answer:
AnnyKZ [126]2 years ago
7 0

Answer:

a) Temperatura, b) Temperature, c)    Constant , d)  None of these , e) Gibbs enthalpy and free energy (G)

Explanation:

a) the expression for ideal gases is PV = nRT

     Temperature

b) The internal energy is E = K T

      Temperature

c)  S = ΔQ/T

In an isolated system ΔQ is zero, entropy  is constant

       Constant

d) all parameters change when changing status

        None of these

e) Gibbs enthalpy and free energy

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Will Mark Brainliest if Correct PLZ!!!!! A bullet is shot at some angle above the horizontal at an initial velocity of 87m/s on
qaws [65]

Answer:

≅50°

Explanation:

We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:

Δx=V₀t+at²/2

And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:

Δx=(V₀cosθ)t+at²/2

Now luckily we are given everything we need to solve (or you found the info before posting here):

  • Δx=760 m
  • V₀=87 m/s
  • t=13.6 s
  • a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!

With that we can plug the values in to get:

760=(87)(cos\theta )(13.6)+\frac{(0)(13.6^{2}) }{2}

760=(1183.2)(cos\theta)

cos\theta=\frac{760}{1183.2}

\theta=cos^{-1}(\frac{760}{1183.2})\approx50^{o}

3 0
3 years ago
A worker assigned to the restoration of the Washington Monument is checking the condition of the stone at the very top of the mo
Mnenie [13.5K]

Answer:

Your answer is: K.E = 8.3 J

Explanation:

If the height (h) = 169.2 meters (m) and the mass (m) is 0.005 kilograms (kg) the total energy will be kinetic energy which is equal to the potential energy.

K.E = P.E and also P.E equals to mgh

Then you substitute all the parameters into the formula  ↓

P.E = 0.005 × 9.81 × 169.2

P.E = 8.2908 J

So your answer is 8.2908 but if you round it is K.E = 8.3

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3 years ago
And event or situation that causes a strain on your body or mind is called a help asap
faltersainse [42]
I'm quite certain the answer is "stress".
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Dan is 50 years old, is beginning to look back at the successes and failures in his life, and hopes to use this reflection to de
Mice21 [21]

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3 0
2 years ago
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Light with an intensity of 1 kW/m2 falls normally on a surface and is completely absorbed. The radiation pressure is
kobusy [5.1K]

Answer:

The radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

Explanation:

Given;

intensity of light, I = 1 kW/m²

The radiation pressure of light is given as;

Radiation \ Pressure = \frac{Flux \ density}{Speed \ of \ light}

I kW = 1000 J/s

The energy flux density = 1000 J/m².s

The speed of light = 3 x 10⁸ m/s

Thus, the radiation pressure of the light is calculated as;

Radiation \ pressure = \frac{1000}{3*10^{8}} \\\\Radiation \ pressure =3.33*10^{-6} \ Pa

Therefore, the radiation pressure of the light is 3.33 x 10⁻⁶ Pa.

6 0
3 years ago
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