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swat32
3 years ago
11

Suppose nuclear physicists discovered a fission process that always produced two smaller nuclei and only one neutron. Would this

eliminate the danger of meltdown in a nuclear power plant? Why or Why not- ? Explain.
(Thanks to anyone who helps :)
Physics
1 answer:
Mila [183]3 years ago
6 0
Running that process in a power plant would not only eliminate the
danger of a core meltdown, it would also pretty much eliminate the
possibility of getting any nuclear power out of the plant.

The understanding behind your question is correct . . . Neutrons coming
out of one fission go on to get absorbed in other nuclei, and cause the
other ones to fiss.  BUT . . . NOT every free neutron whizzing around in
the core material gets absorbed.  Some of them enter another nucleus
and immediately get spit out.  And some of them never get near another
nucleus at all.  Of all the neutrons produced in one fission, only some
percentage go on to stimulate another one.  So if each fission produced
only one neutron, then the whole process would completely run out of gas
in some amount if time, as the number of free neutrons dwindled and shrank.

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four students push carts filled with sports equipment across the gym. Each student pushes with the same amount of force. which c
anastassius [24]

The cart that has the smallest mass of sports equipment in it
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                      Force = (mass) x (acceleration).

If several objects have the same force acting on them, then the one
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6 0
3 years ago
Read 2 more answers
PLEASE HELP ME WITH THIS
Kitty [74]

Answer:

i think it might be c but i know its not A or B and im

on the fence for D so the best answer would be C

Explanation:

3 0
3 years ago
A positive point charge Q1 = 2.5 x 10-5 C is fixed at the origin of coordinates, and a negative point charge Q2 = -5.0 x 10-6 C
mario62 [17]

Answer:

3.62 m  and - 1.4 m

Explanation:

Consider a location towards the positive side of x-axis beyond the location of charge Q₂

x = distance of the location from charge Q₂

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

\frac{kQ_{1}}{(2 + x)^{2}}= \frac{kQ_{2}}{x^{2}}

\frac{2.5\times 10^{-5}}{(2 + x)^{2}}= \frac{5 \times 10^{-6}}{x^{2}}

x = 1.62 m

So location is 2 + 1.62 = 3.62 m

Consider a location towards the negative side of x-axis beyond the location of charge Q₁

x = distance of the location from charge Q₁

d = distance between the two charges = 2 m

For the electric field to be zero at the location

E₁ = Electric field by charge Q₁ at the location = E₂ = Electric field by charge Q₂ at the location

\frac{kQ_{1}}{(x)^{2}}= \frac{kQ_{2}}{ (2 + x)^{2}}

\frac{2.5\times 10^{-5}}{(x)^{2}}= \frac{5 \times 10^{-6}}{(2+x)^{2}}

x = - 1.4 m

6 0
4 years ago
Calculate the frequency of each of the following wave lengths of electromagnetic radiation. Part A 488.0 nm (wavelength of argon
Drupady [299]

Answer:

A. f=6.1475*10^{14}Hz

B. f=5.9642*10^{14}Hz

Explanation:

The frequency has an inversely proportional relationship with the concept of wavelength, the greater the wavelength, the lower the frequency. For electromagnetic waves, the frequency is equal to the speed of light, divided by the wavelength.

f=\frac{c}{\lambda}

A.

f=\frac{3*10^8\frac{m}{s}}{488*10^{-9}m}\\\\f=6.1475*10^{14}Hz

B.

f=\frac{3*10^8\frac{m}{s}}{503*10^{-9}m}\\\\f=5.9642*10^{14}Hz

7 0
4 years ago
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lakkis [162]

Answer:

1.39m

Explanation:

5 0
2 years ago
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