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barxatty [35]
3 years ago
8

Can I have help? please please please!

Physics
2 answers:
andriy [413]3 years ago
8 0
I don't really know but i looked it i up it it said some thing about rod cells 
MA_775_DIABLO [31]3 years ago
7 0
C is the answer sorry if i am wrong
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efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.
PolarNik [594]

Answer:

Inlet : v_i=0.0646\frac{m}{s}

Outlet:  v_o=0.171\frac{m}{s}

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

\dot{m}=\frac{\dot{V}}{\upsilon}, where \dot{V} represent the flow rate and \upsilon the specific volume at the pressure and temperature given.

A_i=0.5m^2 is the inlet area

P_i=600Kpa pressure at the inlet area

T_i=70C temperature at the inlet area

A_o=1m^2 is the outlet area

P_o=100Kpa pressure at the outlet area

T_o=C temperature at the outlet area

\dot{m}=0.75\frac{kg}{s} represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

\upsilon_i =0.04304\frac{kg}{m^3} and the entropy is h_i=1.0645\frac{KJ}{KgK}=h_o

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy h_o=1.0645\frac{KJ}{KgK}

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}

h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}

Our interest value would be given using interpolation like this:

\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that \dot{m}=\frac{\dot{V}}{\upsilon}, but since \dot{V}=Av we have this:

\dot{m}=\frac{Av}{\upsilon}

If we solve from the velocity v we have this:

v=\frac{\upsilon \dot{m}}{A}   (*)

And now we just need to replace the values into equation (*)

For the inlet case:

v_i=\frac{\upsilon_i \dot{m}}{A_i}=\frac{0.043069\frac{kg}{m^3}(0.75\frac{kg}{s})}{0.5m^2}=0.0646\frac{m}{s}

For the oulet case:

v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}

7 0
3 years ago
Explain why a cow that touches an electric fence experiences a mild shock
Minchanka [31]
The voltage exists between the fence and the ground.  The cow is grounded. The cow is touching the ground, completing the circuit of electricity. <span>When the cow comes into contact with the fence, it becomes an electric ground which sends an electric current into the cow, through the cow, and into the ground. The pain experienced from the shock is due to the current that flows through the cow.</span>
8 0
3 years ago
Gary saved dimes and nickels at a ratio of 5:7. Then, he saved 20% more dimes and 3 times the original number of nickels. If he
TiliK225 [7]

Answer:

i think 53 or 56

Explanation:

cause i cross multiplied and got that

8 0
3 years ago
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This exists in the space surrounding a charged particle and exerts a force on other charged particles.
Zinaida [17]

Gravitational field exists in the space surrounding a charged particle and exerts a force on other charged particles. Gravitational waves are ripples of waves travelling outward from the source. The more massive the orbit of two bodies, the more it emits gravitational wave. And everything around it that is near within the wave experiences a ‘pull’ toward the orbiting bodies.

4 0
3 years ago
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What amount of heat is required to raise the temperature of 25 grams of copper to cause a 15ºC change? The specific heat of copp
Lina20 [59]

The amount of heat required is B) 150 J

Explanation:

The amount of heat energy required to increase the temperature of a substance is given by the equation:

Q=mC\Delta T

where:

m is the mass of the substance

C is the specific heat capacity of the substance

\Delta T is the change in temperature of the substance

For the sample of copper in this problem, we have:

m = 25 g (mass)

C = 0.39 J/gºC (specific heat capacity of copper)

\Delta T = 15^{\circ}C (change in temperature)

Substituting, we find:

Q=(25)(0.39)(15)=146 J

So, the closest answer is B) 150 J.

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

3 0
3 years ago
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