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hammer [34]
4 years ago
12

A reactor operating at 1 MW is scrammed by instantaneous insertion of $5.00 of negative reactivity. Approximately how long does

it take the power to drop to 1 milliwatt?
Physics
1 answer:
Novosadov [1.4K]4 years ago
4 0

Answer:

time is 3333.33 min or 55.55 hr

Explanation:

given data

reactor operating = 1 MW

negative reactivity =  $5

power = 1 miliwatt

to find out

how long does it take

solution

we know here power coefficient that is

power coefficient = \frac{10^{6} }{10^{6} }

power coefficient = 1

so time required to reach power is

power =  reactivity × time / power coefficient + reactor operating

1 × 10^{-3} = -5 t / 1  +  1 × 10^{6}

5t =  10^{6} - 10^{-3}

t = 199999.99 sec

so time is 3333.33 min or 55.55 hr

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How to find velocity right before impact?
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4 years ago
A dolphin swims 1.85 km/h how far has the dolphin traveled after 0.60 h
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6 0
3 years ago
A charge of 8.0 pc is distributed uniformly on a spherical surface (radius = 2.0 cm), and a second charge of â3.0 pc is distribu
loris [4]
According to Gauss' law, the electric field outside a spherical surface uniformly charged is equal to the electric field if the whole charge were concentrated at the center of the sphere.

Therefore, when you are outside two spheres, the electric field will be the overlapping of the two electric fields:
E(r > r₂ > r₁) = k · q₁/r² + k  · q₂/r² = k · (q₁ + q₂) / r²
where:
k = 9×10⁹ N·m²/C²

We have to transform our data into the correct units of measurement:
q₁ = 8.0 pC = 8.0×10⁻¹² C
q₂ = 3.0 pC = 3.0×10<span>⁻¹² C
</span><span>r = 5.0 cm = 0.05 m

Now, we can apply the formula:
</span><span>E(r) = k · (q₁ + q₂) / r²
      = </span>9×10⁹ · (8.0×10⁻¹² + 3.0×10⁻¹²) / (0.05)²
      = 39.6 N/C

Hence, <span>the magnitude of the electric field 5.0 cm from the center of the two surfaces is E = 39.6 N/C</span>
4 0
4 years ago
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