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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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On earth which force is 10 to slow an object down
ololo11 [35]

Answer:

gravity

Explanation:

6 0
3 years ago
How significant is air resistance for projectile motion?
miv72 [106K]

Answer:

Air resistance have some significant role in projectile motion if the motion lasts for some time.

Explanation:

  • Air resistance or air drag seems to be important in daily actvities and games like baseball.
  • The trajectory of the projectile with or without air resistance or air drag is totally different.
  • When we neglect air drag, the only acting force is gravity against the motion so the maximum height and range are suppose Hmax and R.
  • Now, when we consider air drag, it is important to notice that there are two forces against the motion of the ball and along the direction of gravity. It seems that both maximum height and range are lesser Hmax'< Hmax and R'<R.
8 0
2 years ago
A 15 m ladder with a mass of 51 kg is leaning against a frictionless wall, which makes an angle of 60 degrees with the horizonta
timofeeve [1]

Answer:

Explanation:

Given that, .

Mass of ladder is 51kg

Then, it weight is

WL = mg = 51 × 9.81 = 500.31N

This weight will act at the midpoint of the ladder

Length of ladder is 15m

The ladder makes an angle 55°C with the horizontal

An object whose mass is 81kg is at 4m from the bottom of the ladder

Then, weight of object

Wo = mg = 81 × 9.81 = 794.61 N

Using newton second law

Check attachment

Ng is normal force on the ground

Ff is the horizontal frictional force

Nw Is the normal force on the wall

ΣFy = 0

Ng = Wo + WL

Ng = 794.61 + 500.31

Ng = 1294.92 N

Also

ΣFx = 0

Ff — Nw = 0

Then,

Ff = Nw

Now taking moment about point A.

Check attachment

using the principle of equilibrium

Sum of clockwise moment equals to sum of anti-clockwise moment

Also note that the Normal force on the wall is not perpendicular to the ladder, so we will resolve that and also the weights of ladder and weight of object

Clockwise = Anticlockwise

Wo•Cos60 × 4 + WL•Cos60 × 7.5 = Nw•Sin60 × 15

794.61Cos60 × 4 + 500.31Cos60 × 7.5 = Nw × Sin60 × 15

1589.22 + 1876.163 = 12.99•Nw

3465.383 = 12.99•Nw

Nw = 3465.383 / 12.99

Nw = 266.77 N

Since, Nw = Ff

Then, Ff = 266.77N

the horizontal force exerted by the ground on the ladder is 266.77 N

8 0
3 years ago
Read 2 more answers
(8 points) Air is used as the working fluid in a simple ideal Brayton cycle that has a pressure ratio of 12, a compressor inlet
trapecia [35]

Answer:

a ) \dot m = 351.49 kg/s

b)  \dot m_{actual} = 1046.15 kg/s

Explanation:

given data:

pressure ration rp = 12

inlet temperature = 300 K

TURBINE inlet temperature  = 1000 K

AT the end of isentropic process (compression) temperature is

\frac{T_2'}{T_1} = rp ^{\frac{\gamma -1}{\gamma}}

\frac{T_2'}{300} = 12^{\frac{1.4 -1}{1.4}}

T_2' = 610.181 K

AT the end of isentropic process (expansion) temperature is

\frac{T_3}{T_4'} = rp ^{\frac{\gamma -1}{\gamma}}

\frac{1000'}{T_4'} = 12^{\frac{1.4 -1}{1.4}}

T_4' = 491.66 K

isentropic work is given as

w(compressor) = CP (T_2' -T_1)

w = 1.005(610.18 - 300)

w = 311.73 kJ/kg

w(turbine) = 1.005( 1000 - 491.66)

w(turbine) = 510.88 kJ/kg

a) mass flow rate for isentropic process is given as

\dot m = \frac{70000}{510.88 - 311.73}

\dot m = 351.49 kg/s

b) actual mass flow rate uis given as

\dot m_{actual} = \frac{70000}{51.088\times 0.85 - \frac{311.73}{0.85}}

\dot m_{actual} = 1046.15 kg/s

6 0
3 years ago
Tarzan swings on a 31.0 m long vine initially inclined at an angle of 36.0◦ with the vertical. The acceleration of gravity if 9.
Gennadij [26K]

Answer:

v=10.777m/s

Explanation:

Tarzan swing can be thought of as change in potential energy by going from higher location We solve for height of beginning of the swing by using simple cosine equation:

So

31Cos36=25.08\\E_{potential}=mgh\\

ΔE=mg(h₂-h₁)

=m*9.81(31-25.08)\\

The potential energy of Tarzan initial position is converted into kinetic energy of his swing.By using kinetic equation

E_{kinectic}=P_{potential}\\1/2mv^{2}=m*9.81(31.0-25.08)\\(1/2)v^{2}=9.81(31.0-25.08)\\0.5v^{2}=58.07\\v^{2}=58.07/0.5\\v=\sqrt{58.07/0.5}\\ v=10.777m/s    

8 0
3 years ago
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