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erica [24]
3 years ago
15

In a nuclear power plant, the nuclear reaction is kept from going critical by keeping the rate of reaction safe.

Physics
1 answer:
Juli2301 [7.4K]3 years ago
6 0

Answer:

B. The control rods absorb excess neutrons, keeping them from causing too many uranium atoms to split too quickly

You might be interested in
Does the planet exert a torque on the meteoroid with respect to the center of mass of the planet? Why or why not?
ra1l [238]

Answer:

yes

Explanation:

because the planet exerts a centripetal force on the meteorold

4 0
3 years ago
A circular parallel plate capacitor is constructed with a radius of 0.52 mm, a plate separation of 0.013 mm, and filled with an
olya-2409 [2.1K]

Answer:

289282

Explanation:

r = Radius of plate = 0.52 mm

d = Plate separation = 0.013 mm

A = Area = \pi r^2

V = Potential applied = 2 mV

k = Dielectric constant = 40

\epsilon_0 = Electric constant = 8.854\times 10^{-12}\ \text{F/m}

Capacitance is given by

C=\dfrac{k\epsilon_0A}{d}

Charge is given by

Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0AV}{d}\\\Rightarrow Q=\dfrac{40\times 8.854\times 10^{-12}\times\pi \times (0.52\times 10^{-3})^2\times 2\times 10^{-3}}{0.013\times 10^{-3}}\\\Rightarrow Q=4.6285\times 10^{-14}\ \text{C}

Number of electron is given by

n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{4.6285\times 10^{-14}}{1.6\times10^{-19}}\\\Rightarrow n=289281.25\ \text{electrons}

The number of charge carriers that will accumulate on this capacitor is approximately 289282.

6 0
3 years ago
Se deja caer una moneda desde cierta altura. Si se desprecian los efectos del aire, ¿cómo varía la fuerza neta sobre la moneda a
forsale [732]

Answer:

Ok, primero pensemos en una situación normal.

La moneda comienza a caer, pero la moneda esta inmersa en una sustancia, el aire. El aire comienza a aplicar una resistencia al movimiento de la moneda, y esta resistencia incremente a medida que la velocidad de la moneda incremente. Llega un punto en el que esta nueva fuerza es igual a la fuerza gravitatoria, y en sentido opuesto, lo que causa que la fuerza neta sea 0, y que la moneda caiga a velocidad constante hasta que esta impacta con el suelo.

Ahora, en este caso tenemos que ignorar los efectos del aire, entonces no hay ninguna fuerza que se oponga a la fuerza gravitatoria, entonces la fuerza neta no cambia a medida que cae (La fuerza neta cambia cuando la moneda impacta el suelo).

También se puede analizar el caso en el que, como la fuerza gravitatoria decrece con el radio al cuadrado, a medida que la moneda cae, la fuerza gravitatoria incrementa. El tema es que en para estas dimensiones, ese cambio en la fuerza gravitacional es imperceptible,

3 0
4 years ago
Please help :<
Artemon [7]

Answer: methyl

Explanation:

7 0
3 years ago
A weight is hung from the ceiling of an elevator by a massless string. Under which circumstances will the tension in the cord be
Oduvanchick [21]
When the elevator is going up (assuming the elevator is acceleration)

When the elevator is accelerating downwards, the total gravitational force would be larger.

If the elevator is accelerating upwards, then the gravitation force would be smaller, thus the tension in the string would be smaller.
4 0
3 years ago
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