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Ksenya-84 [330]
3 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]3 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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The length of a simple pendulum is 0.66 m, the pendulum bob has a mass of 310 grams, and it is released at an angle of 12 degree
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A) the periodic time is given by the equation;
 T= 2π√(L/g)
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T = 2 × 3.14 √ (0.66/9.81)
   = 6.28 × √0.0673
    = 1.6289 Seconds
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b)  The vertical distance, the height is given by
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 h = 0.65 m
Vertical fall at the lowest point = 0.66 - 0.65 = 0.01 m
Applying conservation of energy
energy lost (MgΔh) = KE gained (1/2mv²)
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  v² = 2gΔh = 2×9.81 × 0.01 
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v = 0.443 m/s

c) total energy = KE + GPE = KE when GPE is equal to zero (at the lowest point possible)
Thus total energy is equal to;
E = 1/2mv²
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Read 2 more answers
You have a spring that stretches 0.070 m when a 0.10-kg block is attached to and hangs from it at position y0. Imagine that you
olga nikolaevna [1]

Answer:

a) Δy = 0.144 m

b) W = 0.145 J

c) Us = 0.32 J

d) ymax = 0.144 m

Explanation:

a) First let's find the spring constant using Hooke's Law

F = k*Δy   ⇒  k = F/Δy

where

F = m*g = 0.1 kg*9.81 m/s² = 0.981 N

and  Δy = 0.07 m. Hence

k = 0.981 N/0.07 m = 14.014 N/m ≈ 14 N/m

In order to find the position of the block when we let it go, we need to find the force that caused this expansion in the spring, we know that the reading of the scale was 3 N and this reading includes the force we want to find and the weight of the block, therefore:

f = 3 N - F = 3 N - 0.981 N = 2.019 N

Now that we have found the force we can use Hooke's Law in order to find the position of the block

f = k*Δy   ⇒   Δy = f/k

⇒   Δy = 2.019 N/14 N/m

⇒   Δy = 0.144 m

b) First, notice that there are two kind of potential energy: the potential energy in the spring and the potential energy due to the gravitational field:

W = ΔU

W = ΔUs + ΔUg

W = (Usf - Usi) + (Ugf - Ugi)

Notice that

Us = 0.5*k*y²

where

yf = 0.07 m + 0.144 m = 0.214 m  and

yi = 0.07 m

and we will take the zero level to be the equilibrium position where the block was hanging at rest. Hence

W = 0.5*k*(yf² - yi²) + m*g*(0 - Δy)

⇒ W = 0.5*14 N/m*((0.214 m)² - (0.07 m)²) + (0.1 kg)*(9.81 m/s²)*(0 - 0.144 m)

⇒ W = 0.145 J

c) When we let the block go the spring was stretched by

y = 0.07 m + 0.144 m = 0.214 m

Therefore:

Us = 0.5*k*y²

⇒ Us = 0.5*14 N/m*(0.214 m)²

⇒ Us = 0.32 J

d) Because the position that we pulled the block to it is considered as the amplitude for the vibrational motion that will happen after we release the block, then the maximum height the particle will reach above the equilibrium position is

ymax = Δy = 0.144 m

 

3 0
4 years ago
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