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lozanna [386]
3 years ago
8

A nuclear fission reactor produces electricity. What is the role of the fuel, the control rods, water, and generator in this pro

cess?
A reactor core full of water is heated to steam by the fission reaction that occurs inside control rods full of radioactive fuel pellets. The generator collects the emitted particles as the radioactive nuclei collide and transfers them to the power grid.

Fuel rods heat a pool of water to steam as heat from the fission reaction of the yellowcake inside is transferred to the water. The control rods are used to control the rate of electricity leaving the generator as it converts mechanical energy from a steam-powered turbine into electrical energy.

A generator is used to activate the control rods which contain radioactive isotopes. Once activated, these isotopes begin a nuclear fission chain reaction. Water in a cooling tank keeps the rate of reaction under control as electrons emitted from the reaction are fed through wires to homes and businesses.

The radioactive isotopes used as fuel provide the heat energy to create steam from water through a nuclear fission chain reaction. The control rods are used to regulate the rate of this fission reaction. The generator converts mechanical energy from a steam-powered turbine into electricity.
Physics
2 answers:
77julia77 [94]3 years ago
8 0

Answer:Correct Answer:  

Correct The radioactive isotopes used as fuel provide the heat energy to create steam from water through a nuclear fission chain reaction. The control rods are used to regulate the rate of this fission reaction. The generator converts mechanical energy from a steam-powered turbine into electricity.



Explanation: VERIFIED

natima [27]3 years ago
7 0
The appropriate response is the third one. A generator is utilized to enact the control poles which contain radioactive isotopes. Once initiated, these isotopes start an atomic splitting chain response. Water in a cooling tank monitors the rate of response as electrons radiated from the response are encouraged through wires to homes and organizations.
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A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon is holding a rock. The total
Lynna [10]

Explanation:

Given Data

Total mass=93.5 kg

Rock mass=0.310 kg

Initially wagon speed=0.540 m/s

rock speed=16.5 m/s

To Find

The speed of the wagon

Solution

As the wagon rolls, momentum is given as

P=mv

where

m is mass

v is speed

put the values

P=93.5kg × 0.540 m/s

P =50.49 kg×m/s

Now we have to find the momentum of rock

momentum of rock = mv

momentum of rock = (0.310kg)×(16.5 m/s)

momentum of rock =5.115 kg×m/s  

From the conservation of momentum we can find the wagons momentum So

wagon momentum=50.49 -5.115 = 45.375 kg×m/s  

Speed of wagon = wagon momentum/(total mass-rock mass)

Speed of wagon=45.375/(93.5-0.310)

Speed of wagon= 0.487 m/s

Throwing rock backward,

momentum of wagon = 50.49+5.115 = 55.605  kg×m/s

Speed of wagon = wagon momentum/(total mass-rock mass)

speed of wagon = 55.605  kg×m/s/(93.5kg-0.310kg)

speed of wagon= 0.5967 m/s

7 0
3 years ago
Select the correct answer.
Simora [160]

Answer:

A. 2.36 Newtons

Explanation:

F = GmM/d²

F = 6.673 x 10⁻¹¹(1)(5.98 x 10²⁴) / (1.3 x 10⁷)²

F = 2.36121...

Very poor question design.

  mass of box... 1 significant digit

        distance... 2 significant digits

mass of earth... 3 significant digits

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7.For the following questions, use a periodic table and your atomic calculations to find the unknown information about each isot
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4 years ago
A closely wound, circular coil with a diameter of 4.30 cm has 470 turns and carries a current of 0.460 A .
Nadusha1986 [10]

Hi there!

a)
Let's use Biot-Savart's law to derive an expression for the magnetic field produced by ONE loop.

dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}

dB = Differential Magnetic field element

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

R = radius of loop (2.15 cm = 0.0215 m)

i = Current in loop (0.460 A)

For a circular coil, the radius vector and the differential length vector are ALWAYS perpendicular. So, for their cross-product, since sin(90) = 1, we can disregard it.

dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{r^2}

Now, let's write the integral, replacing 'dl' with 'ds' for an arc length:
B = \int \frac{\mu_0}{4\pi} \frac{ids}{R^2}

Taking out constants from the integral:
B =\frac{\mu_0 i}{4\pi R^2}  \int ds

Since we are integrating around an entire circle, we are integrating from 0 to 2π.

B =\frac{\mu_0 i}{4\pi R^2}  \int\limits^{2\pi R}_0 \, ds

Evaluate:
B =\frac{\mu_0 i}{4\pi R^2}  (2\pi R- 0) = \frac{\mu_0 i}{2R}

Plugging in our givens to solve for the magnetic field strength of one loop:

B = \frac{(4\pi *10^{-7}) (0.460)}{2(0.0215)} = 1.3443 \mu T

Multiply by the number of loops to find the total magnetic field:
B_T = N B = 0.00631 = \boxed{6.318 mT}

b)

Now, we have an additional component of the magnetic field. Let's use Biot-Savart's Law again:
dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}

In this case, we cannot disregard the cross-product. Using the angle between the differential length and radius vector 'θ' (in the diagram), we can represent the cross-product as cosθ. However, this would make integrating difficult. Using a right triangle, we can use the angle formed at the top 'φ', and represent this as sinφ.  

dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} sin\theta}{r^2}

Using the diagram, if 'z' is the point's height from the center:

r = \sqrt{z^2 + R^2 }\\\\sin\phi = \frac{R}{\sqrt{z^2 + R^2}}

Substituting this into our expression:
dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{(\sqrt{z^2 + R^2})^2} }(\frac{R}{\sqrt{z^2 + R^2}})\\\\dB = \frac{\mu_0}{4\pi} \frac{iRd\vec{l}}{(z^2 + R^2)^\frac{3}{2}} }

Now, the only thing that isn't constant is the differential length (replace with ds). We will integrate along the entire circle again:
B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} \int\limits^{2\pi R}_0, ds

Evaluate:
B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} (2\pi R)\\\\B = \frac{\mu_0 iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}

Multiplying by the number of loops:
B_T= \frac{\mu_0 N iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}

Plug in the given values:
B_T= \frac{(4\pi *10^{-7}) (470) (0.460)(0.0215)^2}{2 ((0.095)^2 + (0.0215)^2)^\frac{3}{2}}} \\\\ =  0.00006795 = \boxed{67.952 \mu T}

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Hydrogen, helium, and carbon.
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