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3 years ago

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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A resonant circuit using a 286-nFnF capacitor is to resonate at 18.0 kHzkHz. The air-core inductor is to be a solenoid with clos

lukranit [14]

Answer:

The inductor contains $N = 523962.32$ loops

Explanation:

From the question we are told that

The capacitance of the capacitor is $C = 286nF = 286 * 10^{-9} \ F$

The resonance frequency is $f = 18.0 kHz = 18*10^{3} Hz$

The diameter is $d = 1.1 mm = \frac{1.1 }{1000} = 0.00011 \ m$

The of the air-core inductor is $l = 12 \ m$

The permeability of free space is $\mu_o = 4 \pi *10^{-7} \ T \cdot m/A$

Generally the inductance of this air-core inductor is mathematically represented as

$L = \frac{\mu_o * N^2 \pi d^2}{4 l}$

This inductance can also be mathematically represented as

$L = \frac{1}{w^2}$

Where $w$ is the angular speed mathematically given as

$w = 2 \pi f$

$L = \frac{1}{4 \pi ^2 f^2}$

Now equating the both formulas for inductance

$\frac{\mu_o * N^2 \pi d^2}{4 l} = \frac{1}{4 \pi ^2 f^2}$

making N the subject of the formula

$N = \sqrt{\frac{1}{(2 \pi f)^2} * \frac{4 * l }{\mu_o * \pi d^2 C} }$

$N = \frac{1}{2 \pi f} * \frac{2}{d} * \sqrt{\frac{l}{\pi * \mu_o * C} }$

Substituting value

$N = \frac{1}{ 3.142 * 18*10^{3} * 0.00011 } \sqrt{\frac{12}{ 3.142 * 4 \pi *10^{-7}* 286 *10^{-9}} }$

$N = 523962.32$ loops

4 0

2 years ago

If a system has 475 kcal of work done to it, and releases 5.00 × 102 kJ of heat into its surroundings, what is the change in int

Andrews [41]

First, we convert kcal to joules:
1 kcal = 4.184 kJ
475 kcal = 1987.4 kJ

Now, calculating the change in internal energy:

ΔU = Q + W; where Q is the heat supplied to the system and W is the work done on the system.

ΔU = -500 + 1987.4
ΔU = 1487.4 kJ

4 0

3 years ago

Read 2 more answers

Plzz Help worth 22 points!!! which statement about energy transformations is not true?

Finger [1]

I believe B is the answer

5 0

3 years ago

Tortoise and the Hare

Tanya [424]

The answer would be D. <span>The tortoise moved at a constant velocity throughout the race; the hare stopped to rest periodically.

Hope this helped. Good luck!</span>

3 0

2 years ago

Read 2 more answers

In the Biomedical and Physical Sciences building at MSU there are 135 steps from the ground floor to the sixth floor. Each step

satela [25.4K]

Answer:

W = 16.5 Kj

P = 49.9 Watt

E = 16471

Explanation:

m = 73.5kg

t = 5min 30sec = (5×60) + 30 = 330sec

each step = 16.6cm = 0.166m

h = 135×0.166 = 22.41 m

g = 10 m/s²

(i) W = F × s = W × h = mgh

W = 73.5×10×22.41 = 16471.35

W = 16.5 Kj

(ii) Power = workdone/time

P = 16471.35/330

P = 49.9 Watt

(iii) The energy burnt in this process = 16471

4 0

3 years ago