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3 years ago

Generation of electricity in coal-burning power plants and nuclear power plants both involve _______.

Physics

1 answer:

kolbaska11 [484]3 years ago

5 0

Answer:

Heating water to produce steam which drives a turbine

Explanation:

Generation of electricity in coal-burning power plants and nuclear power plants both involve heating water to produce steam which drives a turbine.

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A non uniform rod has mass

Doss [256]

Answer:

$r_{cm} = L/3$

Explanation:

Mass: M, Length: L.

$\sigma (x) = b(L-x)$

The formula that gives center of mass is

$\vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + ...}{m_1 + m_2 + ...} = \frac{\Sigma m_i \vec{r}_i}{\Sigma m_i}$

In the case of a non-uniform mass density, this formula converts to

$\vec{r}_{cm} = \frac{\int\limits^L_0 {x\sigma(x)} \, dx }{\int\limits^L_0 {\sigma(x)} \, dx }$

where the denominator is the total mass and the nominator is the mass times position of each point on the rod.

We have to integrate the mass density over the total rod in order to find the total mass. Likewise, we have to integrate the center of mass of each point (xσ(x)) over the total rod. And if we divide the integrated center of mass to the total mass, we find the center of mass of the rod:

$\vec{r}_{cm} = \frac{\int\limits^L_0 {x\sigma(x)} \, dx }{\int\limits^L_0 {\sigma(x)} \, dx } = \frac{\int\limits^L_0 {xb(L-x)} \, dx }{\int\limits^L_0 {b(L-x)} \, dx } = \frac{b\int\limits^L_0{(xL - x^2)} \, dx }{b\int\limits^L_0 {(L-x)} \, dx } = \frac{\frac{x^2L}{2} - \frac{x^3}{3}}{Lx - \frac{x^2}{2}}\left \{ {{x=L} \atop {x=0}} \right.$

Here x's are cancelled. Otherwise, the denominator would be zero.

$r_{cm} = \frac{\frac{xL}{2}-\frac{x^2}{3}}{L-\frac{x}{2}}\left \{ {{x=L} \atop {x=0}} \right. = \frac{\frac{L^2}{2}-\frac{L^2}{3}}{L-\frac{L}{2}} = \frac{\frac{L^2}{6}}{\frac{L}{2}} = \frac{L}{3}$

8 0

3 years ago

Air is compressed from 14.7 psia and 60°F to a pressure of 150 psia while being cooled at a rate of 10 Btu/lbm by circulating wa

iren2701 [21]

Answer:

A. 6.36 lbm/s

b. $T_2=341\textdegree F$

Explanation:

a. Given the following information;

#Compressor inlet:

Air pressure, $P_1=14.7psia,T_1=60\textdegree F$

#Compressor outlet:

$Air \ Pressure, P_2=150psia\\\\Air \ volume \ flow \ rate, \dot V_1=5000ft^3/min$

#Cooling rate,

$q_{out}=10Btu/lbm, \dot W=700hp$

# From table A-1E

Gas constant of air $R=0.3704\ psia.ft^3/lbm.R$

Specific enthalpy at $P_1=520R-h_1=124.27Btu/lbm$

Using the mass balance:

$\dot m_{in}-\dot m_{out}=\bigtriangleup \dot m_{sys}=0.0\\\\\dot m_{in}=\dot m_{out}\\\\v_1=\frac{RT_1}{P_1}\\\\v_1=\frac{0.3704\times520}{14.7}\\\\\\v_1=13.1026ft^3/lbm\\\\\dot m=\frac{5000/60}{13.1025}\\\\\dot m=6.36\ lbm/s$

Hence, the mass flow rate of the air is 6.36lbm/s

b The specific enthalpy at the exit is defined as the energy balance on the system:

$\dot m_{in}-\dot m_{out}=\bigtriangleup \dot m_{sys}=0.0\\\\\dot m_{in}=\dot m_{out}\\\\\dot W_{in}+ \dot mh_1=\dot Q_{out}+\dot m_2\\\\h_2=\frac{\dot W_{in}-\dot Q_{out}}{\dot m}+h_1\\\\h_2=\frac{700\times 0.7068-10\times 6.36}{6.36}+124.27\\\\h_2=192.06\ Btu/lbm\\\\#at \ h_2=182.06Btu/lbm, T_2=801R=341\textdegree F$

Hence, the temperature at the compressor exit $T_2=341\textdegree F$

5 0

4 years ago

If the bar magnet is flipped over and the south pole is brought near the hanging ball, the ball will be?

disa [49]

The ball may attracted to the magnet.

<h3>How can we understand that the hanging ball will be attracted to the magnet or not?</h3>

From the question, we understand that the ball is attracted by the north pole of the bar magnet, then the bar magnet flipped over and the south pole is brought near the hanging ball.
As we know, in this type of experiments of bar magnet most of the times the ball is made out of steel.
Steel is a magnetic material.
Magnetic materials gets attracted to the magnet at both the North and South pole.
This can be compared to how neutral objects also gets attracted to the positively and negatively charged rods through the Polarization force.

So, If the bar magnet is flipped over and the south pole is brought near the hanging ball, The ball will be attracted to the magnet.

Learn more about the bar magnet:

brainly.com/question/27943723

#SPJ4

7 0

2 years ago

What if someone refers to a crystal's " habit" what are they referring to?

Alenkinab [10]

It's the external shape of the crystal. You can determine how developed the crystals are and it'll help identify what species of mineral it is. Hope that helped

7 0

3 years ago

A metal cube measures 15 cm on a side and has a density of 6920 kg/m^3. Find the mass in grams.

soldi70 [24.7K]

Answer:

Mass of the cube is 23350 grams.

Explanation:

It is given that,

Side of cube, a = 15 cm = 0.15 m

Density of the cube, $d=6920\ kg/m^3$