All categories

3 years ago

32. For each of the listed parts of a power plant, make a selection to indicate in what

Physics

1 answer:

k0ka [10]3 years ago

7 0

Answer: a. boiler - I. part of a coal-fired power plant.

b. combustion chamber - I. part of a coal-fired power plant.

c. condenser - III. part of a coal-fired power plant and part of a nuclear power plant.

d. control rod - I. part of a coal-fired power plant.

e. generator- III. part of a coal-fired power plant and part of a nuclear power plant.

f. turbine - III. part of a coal-fired power plant and part of a nuclear power plant.

Explanation:

a. Boiler: It is used in the coal power plant to heat the water used for producing electricity. The boiler keeps the temperature of water or steam at a desired level.

b. Combustion chamber: It is used for combustion of coal and using thermal energy for electricity generation.

c. Condenser; It is used in both nuclear and coal power plants. It condense and receive from the turbine.

d. Control rod: It is used in the nuclear power plant for controlling the nuclear fission and fusion reactions of heavy metals.

e. Generator: It converts the thermal energy into electrical energy.

f. Turbine: The turbine blades move with the effect of steam or any form of energy and the energy so produce is used for generating electricity in generator.

You might be interested in

In a carnival booth, you can win a stuffed giraffe if you toss a quarter into a small dish. the dish is on a shelf above the poi

Georgia [21]

components of the speed of the coin is given as

$v_x = v cos60$

$v_x = 6.4 cos60 = 3.2 m/s$

$v_y = vsin60$

$v_y = 6.4 sin60 = 5.54 m/s$

now the time taken by the coin to reach the plate is given by

$t = \frac{\delta x}{v_x}$

$t = \frac{2.1}{3.2}$

$t = 0.656 s$

now in order to find the height

$h = vy * t + \frac{1}{2} at^2$

$h = 5.54 * 0.656 - \frac{1}{2}*9.8*(0.656)^2$

$h = 1.52 m$

so it is placed at 1.52 m height

3 0

3 years ago

What is the greenhouse effect?

Zanzabum

A on that problem :)[email protected]

6 0

3 years ago

Read 2 more answers

An astronaut exploring a distant solar system lands on an unnamed planet with a radius of 2530 km. When the astronaut jumps upwa

Natali [406]

Answer:

1.38*10^18 kg

Explanation:

According to the Newton's law of universal gravitation:

$F=G*\frac{m_a*m_p}{r^2}$

where:

G= Gravitational constant (6.674×10−11 N · (m/kg)2)

ma= mass of the astronaut

mp= mass of the planet

$F=m_a.a\\(v_f )^2=(v_o)^2+2.a.\Delta y\\\\a=\frac{(v_f)^2-(v_o)^2}{2.\Delta y}\\\\a=\frac{(0)^2-(4.29m/s)^2}{2.0.64m}=14.38m/s^2\\\\F=m_a*14.38m/s^2$

so:

$m_a*14.38m/s^2=(6.674*10^{-11}N.(m/kg)^2)*\frac{m_a.m_p}{(2.530*10^3m)^2}\\m_p=\frac{14.38m/s^2(2.530*10^3m)^2}{(6.674*10^{-11}N.(m/kg)^2)}\\\\m_p=1.38*10^{18}kg$