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Crazy boy [7]
3 years ago
10

Will give brainliest! 50 points!!!

Physics
1 answer:
tatiyna3 years ago
6 0

Answer:

C

Im not totally sure but that's what my science teacher taught me sorry if is wrong

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How long does it take a man to travel 6 km if his speed is 3km/h?
Ymorist [56]

why did my answer get deleted??

oh yeah i put a link on there- oopsies.

I wont this time!

I got 30!

3 0
2 years ago
How does unequal solar heating lead to the Gulf Stream?
adoni [48]
Http://earthguide.ucsd.edu/virtualmuseum/virtualmuseum/OriginofGulfStream.shtml this website might help u find ur answer
8 0
3 years ago
Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500◦C, and 80 m/s, and the exit
Cerrena [4.2K]

Answer:

a) ΔEC=-23.4kW

b)W=12106.2kW

c)A=0.01297m^2

Explanation:

A)

The kinetic energy is defined as:

\frac{m*vel^2}{2} (vel is the velocity, to differentiate with v, specific volume).

The kinetic energy change will be: Δ (\frac{mvel^2}{2})=\frac{m*vel_2^2}{2}-\frac{m*vel_1^2}{2}

Δ (\frac{mvel^2}{2})=\frac{m}{2}*(vel_2^2-vel_1^2)

Where 1 and 2 subscripts mean initial and final state respectively.

Δ(\frac{mvel^2}{2})=\frac{12\frac{kg}{s}}{2}*(50^2-80^2)\frac{m^2}{s^2}=-23400W=-23.4kW

This amount is negative because the steam is losing that energy.

B)

Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).

H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)

We already know the last quantity: \frac{m}{2} *(vel_1^2-vel_2^2)=-Δ (\frac{mvel^2}{2})=23400W

For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that, h_1=h(T=500C,P=4MPa)=3445.3\frac{kJ}{kg}

The exit state is a liquid-vapor mixture, so its enthalpy is:

h_2=h_f+xh_{fg}=289.23+0.92*2366.1=2483.4\frac{kJ}{kg}

Finally, the work can be obtained:

W=12\frac{kg}{s}*(3445.3-2438.4)\frac{kJ}{kg} +23.400kW)=12106.2kW

C) For the area, consider the equation of mass flow:

m=p*vel*A where p is the density, and A the area. The density is the inverse of the specific volume, so m=\frac{vel*A}{v}

The specific volume of the inlet steam can be read also from the steam tables, and its value is: 0.08643\frac{m^3}{kg}, so:

A=\frac{m*v}{vel}=\frac{12\frac{kg}{s}*0.08643\frac{m^3}{kg}}{80\frac{m}{s}}=0.01297m^2

Download pdf
7 0
3 years ago
I'm trying to find the potential energy of a planet using formula G M(sun) x m(planet) / r. I have given G - newton's graviation
ludmilkaskok [199]

Answer:

-5.39\times10^{33} J

Explanation:

Potential energy =

-\frac{GMm}{r}\\=-\frac{6.67\times10^{-11}\times1.99\times10^{30}\times5.97\times10^{24}}{147/1\times10^{9}}\\=-5.39\times10^{33} J

7 0
2 years ago
When a customer in Japan was quoted a price of .39 cents a pound he thought he was being charged .39 a kilogram. He was shocked
Anon25 [30]

Answer: $85.80

Explanation: 100kg is equalvilent to 220 lbs, then you multiply 220 by .39 and then you have your answer :)

4 0
3 years ago
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