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3 years ago

10

The controlled release of nuclear energy in a reactor is accomplished by: allowing the critical mass of fuel to react controllin

g a chain reaction controlling fission without a chain reaction all of the above

1 answer:

amm18123 years ago

3 0

The controlled release of nuclear energy in a reactor is accomplished by controlling fission without a chain reaction.

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Integrate your expressions for dEx and dEy from θ=0 to θ=π. The results will be the x-component and y-component of the electric

Nimfa-mama [501]

Answer:

hello your question is incomplete below is the missing part

Ex = 0

Ey = $\frac{-2kQ}{\pi a^2}$

Explanation:

Attached below is a detailed solution showing the integration of the expression dEx and dEy from ∅ = 0 to ∅ =π

Ex = 0

Ey = $\frac{-2kQ}{\pi a^2}$

6 0

3 years ago

At the equator, near the surface of the Earth, the magnetic field is approximately 50.0 μT northward, and the electric field is

MArishka [77]

Answer:

A) F_g = 8.9278 × 10^(-30)

B) F_e = 1.6 × 10^(-17) N upwards

C) F_m = 6.8 × 10^(-17) N downwards

Explanation:

A) Formula for gravitational force is;

F_g = m_e × g

m_e is mass of electron = 9.11 x 10^(-31) kg

F_g = 9.11 x 10^(-31) × 9.8

F_g = 8.9278 × 10^(-30) N downwards

B) Formula for Electric force is;

F_e = qe

q is charge on electron = 1.6 × 10^(-19) C

E is electric field = 100 N/C

F_e = 1.6 × 10^(-19) × 100

F_e = 1.6 × 10^(-17) N upwards

C) Magnetic force is given by the formula;

F_m = qVB

q is charge on electron = 1.6 × 10^(-19) C

V is velocity given as 8.50 × 10^(6) m/s

B is magnetic field = 50.0 μT = 50 × 10^(-6) T

F_m = 1.6 × 10^(-19) × 8.50 × 10^(6) × 50 × 10^(-6)

F_m = 6.8 × 10^(-17) N downwards

6 0

3 years ago

Discuss how oxygen is used in spacecraft's air supplies in at least 3 paragraphs.

galina1969 [7]

Short-duration spacecraft typically have one backup system and carry their own supply of oxygen. A large portion of the required oxygen is produced on long-duration missions, such as the International Space Station (ISS), which has been in orbit since 1998. Different sources provide the oxygen utilized on the ISS. The water electrolyzer is the primary source of metabolic oxygen. As an alternative to the electrolyzer, oxygen candles (also known as SFOGs) can produce metabolic oxygen. Additionally, oxygen is carried up whenever a cargo ship docks and stored in two tanks on the ISS Airlock. The electrolyzer electrolyzes water to create oxygen by running an electric current through it. Since water is a poor electrical conductor by itself, a little quantity of common salt is dissolved in the water to improve its electrical conductivity. Water is split into hydrogen and oxygen throughout the process.

We must keep in mind that oxygen by itself cannot be inhaled; it must be combined in the proper ratio with nitrogen to make it breathable. Two tanks aboard the ISS are used to store nitrogen, and the cargo ships that travel by from time to time also transport nitrogen cylinders. Through the electrical grid of the station, the solar panels on the station supply the necessary electricity for the oxygen generators. The majority of the required water is transported to the station by cargo supply ships. Condensers, which draw water vapor even from the station's air, ensure that not a drop of water is wasted. Using the proper equipment, water is also recycled from the astronauts' urine.

Through a suitable vent, the hydrogen gas produced during the electrolysis process is released into space. Pressurized tanks at the airlock nodes at the space station are pumped with oxygen when the cargo vehicles arrive there. Pressurized tanks there are also pumped with nitrogen. It goes without saying that the station's atmospheric controls combine the gases in the right amounts for the atmosphere of Earth and then distribute the combination throughout the cabin. The production of oxygen in space is impossible.

3 0

2 years ago

The force between two interacting charges is 9.0 × 10-5 newtons. They are kept 1 meter apart. If the magnitude of one charge is

Oliga [24]

The magnitude of other charge will be 1 × 10⁻² coulomb

The formula of electrostatic force is

Electrostatic force = K q1 q1 / r²

where k is the coulomb's constant whose value is 9 × 10⁹

q1 and a2 are the magnitude of charges

and r is the distance between them

magnitude of the force given to us is 9.0 × 10⁻⁵ newtons

magnitude of one charge = 1.0 × 10⁻⁶ coulomb

Force = K q1 q2 / r²

9.0 × 10⁻⁵ = ( ( 9 × 10⁹ ) × ( 1.0 × 10⁻⁶ ) × q2 ) / 1

9.0 × 10⁻⁵ = 9 × 10³ × q2

10⁻² = q2

Charge on q2 is 1 × 10⁻² coulomb

So the magnitude of the second charge is came out to be 1 × 10⁻² coulomb after applying the formula of electrostatic force.

Learn more about electrostatic force here:

brainly.com/question/17692887

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8 0

2 years ago

A box of mass 20kg is pulled up an inclined plane by a force of 285N. Given that the value of the incline angle is 30 degrees an

docker41 [41]

Given :

Mass of box , m = 250 kg.

Force applied , F = 285 N.

The value of the incline angle is 30°.

the coefficient of dynamic friction is $\mu=0.72$ .

To Find :

The speed with which the box is moving with, assuming it takes 4 seconds to reach the top of the incline.

Solution :

Net force applied in box is :

$F=285 - mgsin\ \theta - \mu mg cos \ \theta\\ \\F=285-mg( sin \ \theta - \mu cos\ \theta)\\\\F=285 - 20\times 10( \dfrac{1}{2}+0.72\times \dfrac{\sqrt{3}}{2})\\\\F=60.29\ N$

Acceleration , $a=\dfrac{F}{m}=\dfrac{60.29}{20}=3.01\ m/s^2$ .

By equation of motion :

$v=u+at\\\\v=0+3.01\times 4\\\\v=12.04\ m/s$

Therefore, the speed of box is 12.04 m/s.

Hence, this is the required solution.

6 0

3 years ago