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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.

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A conical container of radius 6 ft and height 18 ft is filled to a height of 11 ft of a liquid weighing 64.4 lb divided by ft cu

prisoha [69]

Answer:

Hello there, please see explanations for step by step answer.

Explanation:

Radius 6 ft and

Height 18 ft is filled to a height of 11 ft of a liquid weighing 64.4 lb divided by ft cubedlb/ft3.

How much work will it take to pump the contents to the rim.

See attached documents for clear solvings and further step by step explanations

4 0

3 years ago

Two objects, Object A and Object B, need to be identified. Object A's index of refraction is determined to be 1.77, and Object B

Slav-nsk [51]

The correct answer is

C. Light can pass through Object B faster than it can pass through Object A.

In fact, the index of refraction of a material is defined as:

$n=\frac{c}{v}$

where c is the speed of light in vacuum and v is the speed of light in the material. Rearranging the equation, we can write the speed of light in the material as:

$v=\frac{c}{n}$

So we that, the smaller the refractive index n, the greater the speed of light in the material, v. In this problem, object B has lower refractive index than object A, so light travels faster in object B.

4 0

3 years ago

Read 2 more answers

What speed would a fly with a mass of 0.55g need in order to have a kinetic energy of 7.6 •10^4 j?

masya89 [10]

Answer:

16613 m/s

Explanation:

Given that

mass of the fly, m = 0.55 g = 0.55*10^-3 kg

Kinetic Energy of the fly, E = 7.6*10^4 J

Speed of the fly, v = ? m/s

We know that the Kinetic Energy is that energy that an object, in this case, the fly, possesses due to its motion.

The Kinetic Energy, KE of any object is represented by the formula

KE = 1/2 * m * v²

If we substitute the values in the relation, we have,

7.6*10^4 = 1/2 * 0.55*10^-3 * v²

v² = (15.2*10^4) / 0.55*10^-3

v² = 2.76*10^8

v = √2.76*10^8

v = 16613 m/s

Thus, the fly would need a speed of 16.6 km/s in order to have a Kinetic Energy of 7.6*10^4 J

7 0

3 years ago

You have a bowling ball with a mass of 4kg. You throw it with an acceleration of 10 m/s/s. With how much force will it hit the p

Eddi Din [679]

Answer:

<h3>The answer is 40 N</h3>

Explanation:

The force acting on an object can be found by using the formula

<h3>force = mass × acceleration</h3>

From the question

mass = 4 kg

acceleration = 10 m/s²

So we have

force = 4 × 10

We have the final answer as

Hope this helps you

6 0

3 years ago

1500 kg wrecking ball traveling at a speed of 3.5 m/s hits a wall that does not crumble but is pushed back 75 cm. If the wreckin

Rudiy27

Answer:

The size of the force that pushes the wall is 12,250 N.

Explanation:

Given;

mass of the wrecking ball, m = 1500 kg

speed of the wrecking ball, v = 3.5 m/s

distance the ball moved the wall, d = 75 cm = 0.75 m

Apply the principle of work-energy theorem;

Kinetic energy of the wrecking ball = work done by the ball on the wall

¹/₂mv² = F x d

where;

F is the size of the force that pushes the wall

¹/₂mv² = F x d

¹/₂ x 1500 x 3.5² = F x 0.75

9187.5 = 0.75F

F = 9187.5 / 0.75

F = 12,250 N

Therefore, the size of the force that pushes the wall is 12,250 N.

7 0

3 years ago