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sladkih [1.3K]
3 years ago
11

During a heat wave, there is an increased demand for electricity to operate air conditioners. The basic design of a nuclear reac

tor that generates electricity is shown below. What is the best way for the power plant manager to modify the reactor to meet the demand for electricity?
a)lower the control rods, which will absorb more neutrons before they can split atoms
b)raise the control rods so that fewer neutrons are absorbed and more can split uranium atoms
c)remove some of the fuel rods, which will leave more room for additional neutrons to split atoms
d)add additional fuel rods to generate additional neutrons to split uranium atoms
Physics
2 answers:
Lelechka [254]3 years ago
4 0
Given that the design is not shown, for traditional nuclear reactors with control rods, the best way to meet an increased demand for energy would be to raise the control rods so that less neutrons are absorbed.

So, the most reasonable answer to the question is B.
nlexa [21]3 years ago
3 0

Answer:

b)raise the control rods so that fewer neutrons are absorbed and more can split uranium atoms

Explanation:

Here in nuclear reactor we know that high energy neutrons are projected on uranium atom which will further break the uranium into smaller atoms and thus energy is released.

In this whole process the reaction and the released energy is controlled by the help of control rods which will absorb the high speed neutrons and control the energy of the reaction.

Now if energy demand is increased then in this case the energy demand can be fulfilled by generating more energy for which we can remove the control rods so that more number of neutrons will strike with uranium atom and hence more energy will release.

so correct answer will be

b)raise the control rods so that fewer neutrons are absorbed and more can split uranium atoms

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a machine is used to lift crates of oranges has an efficiency of 65%. if the machine provides 1300 newtons through a distance of
mylen [45]

Given data

  efficiency (η) = 65 %

                        = 0.65

  Machine provides (Work done by the machine) = 1300 × 50

                                                                                 = 65000 J

  efficiency (η) = work done by the machine ÷ work supplied to the machine

               0.65 =  65000 ÷ Input work

              Input work = 100000 J

                                = 100 KJ

5 0
3 years ago
Please help me ?????
castortr0y [4]
3 because opposites attract 
6 0
4 years ago
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A magnetic field of 0.276 T exists in the region enclosed by a solenoid that has 517 turns and a diameter of 10.5 cm. Within wha
Advocard [28]

Answer:

The period the field must be reduced to zero is 9.81 x 10⁻⁵ s

Explanation:

Given;

initial value of the magnetic field, B₁ = 0.276 T

number of turns of the solenoid, N = 517 turns

diameter of the solenoid, d = 10.5 cm = 0.105 m

induced emf, = 12.6 kV = 12,600 V

when the field becomes zero, then the final magnetic field value, B₂ = 0

The induced emf is given by Faraday's law;

emf = -\frac{NA\Delta B}{t} \\\\emf = -\frac{NA (B_2 -B_1)}{t} \\\\t = -\frac{NA (B_2 -B_1)}{emf}\\\\t = \frac{NA (B_1 -B_2)}{emf}\\\\where;\\\\t \ is \ the \ time \ when \ B = 0 \ \ (i.e\ B_2 = 0)\\\\A \ is \ the \ area \ of \ the \ coil\\\\A = \frac{\pi d^2}{4} = \frac{\pi (0.105)^2}{4} = 0.00866 \ m^2\\\\t= \frac{(517) \times (0.00866)\times  (0.276 -0)}{12,600}\\\\t = 9.81 \times 10^{-5} \ s

Therefore, the period the field must be reduced to zero is 9.81 x 10⁻⁵ s

4 0
4 years ago
A 3.5 kW drill transfers 5 000 kJ of kinetic energy during 15 seconds of use. What is the percentage efficiency of the drill?
Marizza181 [45]

Answer:

1.04%

Explanation:

Given that,

The power of drill = 3.5 kW = 3500 W

Transferred kinetic energy = 5000 kJ during 15 seconds of use.

We need to find the percentage efficiency of the drill. It can be given by :

\eta=\dfrac{P_o}{P_i}\times 100

Where

Po and Pi are output and input powers.

P_o=\dfrac{5000\times 10^3}{15}\\\\=3.34\times 10^5\ W

So,

\eta=\dfrac{3500}{3.34\times10^{5}}\times100\\\\=1.04\%

So, the percentage efficiency of the drill is 1.04%.

4 0
3 years ago
The dwarf planet Pluto was discovered in 1930. Since that time, which jovian planet has completed a full revolution around the S
Eva8 [605]

Answer:

Explanation:

Uranus takes about 84 years to circle the sun.

Pluto's existance has been known for 92 years. Uranius has completed 1 complete revolution. The numbers are reasonably close together and so Uranius is the answer.

6 0
3 years ago
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