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Ksenya-84 [330]
2 years ago
8

Control rods are used to slow down the reaction in the reactor core when the core becomes too hot. Please select the best answer

from the choices provided T F.
Physics
1 answer:
viktelen [127]2 years ago
5 0

The best answer is true. Yes, control rods are used to slow down the reaction in the reactor core when the core becomes too hot.

<h3>What is a nuclear reactor?</h3>

A nuclear reactor, often known as an atomic pile, is a device that initiates and controls nuclear chain reactions or nuclear fusion processes.

Nuclear reactors are used in nuclear power plants to generate energy as well as in nuclear marine propulsion.

Control rods are employed in nuclear reactors because the processes that occur in a nuclear reactor are very exothermic.

As a result, when these control rods are infused into a nuclear reactor, they absorb neutrons and aid in the management of the chain reaction. As a result, the rate of the chain reaction is brought under control.

As a result, we may infer that the statement control rods are employed to slow down the reaction in the reactor core when it gets too hot is correct.

Hence the best answer is true. Yes, control rods are used to slow down the reaction in the reactor core when the core becomes too hot.

To learn more about the nuclear reactor refer to the link;

brainly.com/question/896706

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Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 24 kg and the larger bottom crate has a m
marusya05 [52]

Answer:

The sum of all forces for the two objects with force of friction F and tension T are:

(i) m₁a₁ = F

(ii) m₂a₂ = T - F

1) no sliding infers: a₁ = a₂= a

The two equations become:

m₂a = T - m₁a

Solving for a:

a = T / (m₁+m₂) = 2.1 m/s²

2) Using equation(i):

F = m₁a = 51.1 N

3) The maximum friction is given by:

F = μsm₁g

Using equation(i) to find a₁ = a₂ = a:

a₁ = μs*g

Using equation(ii)

T = m₁μsg + m₂μsg = (m₁ + m₂)μsg = 851.6 N

4) The kinetic friction is given by: F = μkm₁g

Using equation (i) and the kinetic friction:

a₁ = μkg = 6.1 m/s²

5) Using equation(ii) and the kinetic friction:

m₂a₂ = T - μkm₁g

a₂ = (T - μkm₁g)/m₂ = 12.1 m/s²

4 0
3 years ago
A bike travels at 7.5 m/s along a straight road, whereas a car travels at 10.0 m/s along the same road and in the same direction
Ilya [14]

Answer:

t = 8.3 s

Explanation:

As we know that

velocity of bike = 7.5 m/s

velocity of car is 10 m/s

deceleration of car is 0.75 m/s^2

part a)

velocity of bike with respect to car is given as

v_r = 7.5 - 10 = -2.5 m/s

acceleration of bike with respect to car is given as

a_r = 0 - (-0.75) = 0.75 m/s^2

now the distance of the bike with respect to car is given as

d = v_i t + \frac{1}{2}at^2

5 = (-2.5) t + \frac{1}{2}(0.75)t^2

t = 8.3 s

Part b)

3 0
3 years ago
What happens to a wave when it moves from one medium to another?
Nina [5.8K]
One side of the wave changes speed before the other side, causing the wave to move
3 0
3 years ago
A mass hanged on a spring scale. what is the force exerted by gravity on 700g ?
Ipatiy [6.2K]

Answer:

6.86 N

Explanation:

Applying,

F = mg............... Equation 1

Where F = Force exerted by gravity on the mass, m = mass, g = acceleration due to gravity

Note: The Force exerted by gravity on the mass is thesame as the weight of the body.

From the question,

Given: m = 700 g = (700/1000) = 0.7 kg

Constant: g = 9.8 m/s²

Substitute these values into equation 1

F = 9.8(0.7)

F = 6.86 N

6 0
2 years ago
a 100g ice cube at 0 degrees celsius is placed in 650 grams of water at 25 degrees celsius. When the mixture reaches equillibriu
Artyom0805 [142]

Answer:

The latent heat of fusion of water is 334.88 Joules per gram of water.

Explanation:

Let the latent heat of ice be 'x' J/g

1) Thus heat absorbed by 100 gram of ice to get converted into water equals

Q_1=100\times x

2) heat energy required to raise the temperature of water from 0 to 25 degree Celsius equals

Q_2=100\times 4.186\times 11=4604.6Joules

Thus total energy needed equals Q_1+Q_2=100x+4604.6

3) Heat energy released by the decrease in the temperature of water from 25 to 11 degree Celsius is

Q_3=650\times 4.186\times (25-11)\\\\Q_{3}=38092.6Joules

Now by conservation of energy we have

Q_1+Q_2=Q_3\\\\100x+4604.6=38092.6\\\\\therefore x=\frac{38092.6-4604.6}{100}=334.88J/g

6 0
3 years ago
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