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Ber [7]
1 year ago
7

Which would most likely cause a decrease in the rate of energy production in a fusion nuclear reactor?

Physics
2 answers:
White raven [17]1 year ago
7 0

Answer:

A: a drop in the temperature in the reactor

Explanation:

right on edg

lesya [120]1 year ago
4 0

In a fusion nuclear reactor, a drop in temperature would most likely result in a decline in the rate of energy output.

<h3 /><h3>What is nuclear power?</h3>

The utilization of nuclear reactions to generate energy is known as a nuclear power. Nuclear fission, nuclear decay, and nuclear fusion processes are all sources of nuclear energy.

Nuclear power facilities currently produce the great bulk of the electricity generated by nuclear fission of uranium and plutonium.

The environment must be hot enough for the deuterium and tritium ions' kinetic energies to be sufficient to break through the Coulomb barrier and fuse together.

Hence a decrease in temperature would most likely cause a decrease in the rate of energy production in a fusion nuclear reactor.

To learn more about nuclear power refer to the link;

brainly.com/question/9859575

#SPJ1

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A lunar eclipse can only happen during a(1) new moon.(2) solstice.(3) first quarter moon.(4) full moon.(5) perihelion passage of
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(4) full moon.

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A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behav
gayaneshka [121]

Answer:

a) x = 0.098

b) x = 2.72 m

Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2        (1)

m: mass of the person = 62kg

k: spring constant = ?

v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

v^2=v_o^2+2gy

vo: initial velocity = 0 m/s

g: gravitational acceleration = 9.8 m/s^2

y: height from the person jumps = 20.0m

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}

With this value you can find the spring constant k from the equation (1):

mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

W=F_e\\\\mg=kx

you solve the last expression for x:

x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}

Next, you calculate x from the equation (1):

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m

The net fire is stretched 2.72 m

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How planets look like,and maybe How the planets were created
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