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vitfil [10]
3 years ago
11

What process is used to release energy in nuclear power plants

Physics
2 answers:
Hunter-Best [27]3 years ago
6 0
Fission is the process used to release energy in nuclear power plants (:
Anika [276]3 years ago
5 0
.........Nucler fission.... .
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Which of the following statements is true for a cell placed in beaker containing an isotonic solution?
Korvikt [17]
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The correct answer is <u>C. w</u></span><span><u>ater molecules flow in both directions at the same rate.</u></span>
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3 years ago
A wave moving along a rope has a
Gennadij [26K]

Answer:

V = wavelength * frequency = 1.5 * 5.5 = 8.25 m/s

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the weight of 80 kg of mass on mercury is 296 N and almost identical to the weight of the same mass on Mars but mercury has much
Dmitrij [34]
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What is the difference between potential energy and kinetic energy?
gavmur [86]
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6 0
3 years ago
If the earth's magnetic field has strength 0.50 gauss and makes an angle of 20.0 degrees with the garage floor, calculate the ch
lys-0071 [83]

Answer:

ΔΦ = -3.39*10^-6

Explanation:

Given:-

- The given magnetic field strength B = 0.50 gauss

- The angle between earth magnetic field and garage floor ∅ = 20 °

- The loop is rotated by 90 degree.

- The radius of the coil r = 19 cm

Find:

calculate the change in the magnetic flux δφb, in wb, through one of the loops of the coil during the rotation.

Solution:

- The change on flux ΔΦ occurs due to change in angle θ of earth's magnetic field B and the normal to circular coil.

- The strength of magnetic field B and the are of the loop A remains constant. So we have:

                         Φ = B*A*cos(θ)

                         ΔΦ = B*A*( cos(θ_1) - cos(θ_2) )

- The initial angle θ_1 between the normal to the coil and B was:

                         θ_1  = 90° -  ∅

                         θ_1  = 90° -  20° = 70°

The angle θ_2 after rotation between the normal to the coil and B was:

                         θ_2  =  ∅

                         θ_2  = 20°

- Hence, the change in flux can be calculated:

                        ΔΦ = 0.5*10^-4*π*0.19*( cos(70) - cos(20) )

                        ΔΦ = -3.39*10^-6

                       

8 0
3 years ago
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