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3 years ago

Which is an example of current electricity?

1 answer:

6 0

Examples are, starting a car, turning on a light, cooking on an electric stove, watching tv, shaving with an electric razor, playing video games, using a phone, and charging a cell phone, i found this information on www.electricityforum.com

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Solve this for me please

Rama09 [41]

Power output = V*I=11000*750=8250 kVA= 8250 kW

8 0

4 years ago

A 645-turn coil with a 20.250 m 2 area is spun in Earth’s 5.00×10 −5 T magnetic field, producing a 1.25-V maximum emf. A

Dmitriy789 [7]

Answer:

$\omega = 1.914\ rad/s$

Explanation:

Given,

Number of turns, N = 645 N

Area, A = 20.25 m²

Earth Magnetic field, B = 5 x 10⁻⁵ T

Maximum Emf = 1.25 V.

Angular velocity, ω = ?

Using Induced Emf formula

$\varepsilon = NAB\omega$

$\omega= \dfrac{\varepsilon}{NAB}$

$\omega= \dfrac{1.25}{645\times 20.25\times 5\times 10^{-5}}$

$\omega = 1.914\ rad/s$

Angular velocity of the coil = $\omega = 1.914\ rad/s$

5 0

3 years ago

Which statements describe density? Check all that apply.

HACTEHA [7]

Density is a chemical property of an object.

The density of an object is constant.

Density is a derived unit of measure

7 0

3 years ago

Read 2 more answers

Which expression is equivalent to g−m ÷ gn?

Vinvika [58]

<u>Answer</u>

(g²n - m)/(gm)

<u>Explanation</u>

g - m ÷ gn = g - m/gn

Make the equation have the same denominator

g - m ÷ gn = g - m/gn = (ggn)/gn - m/gn

= (g²n)/gn - m/gm

Since they have the same denominator, we can carry out the subtraction on the numerator and then put them under one denominator.

(g²n)/gn - m/gm = (g²n - m)/(gm)

5 0

3 years ago

Uranium was used to generate electricity in a nuclear reactor of a nuclear power station. The nuclear reactor was 29% efficient

Minchanka [31]

Answer:

$C_{lifetime} = 2,618,017,174\,USD$

Explanation:

The energy contained per second in the fuel is:

$\dot E_{fuel} = \frac{800\times 10^{6}\,W}{0.29}$

$\dot E_{fuel} = 2.758\times 10^{9}\,W$

It is know that fission of a gram of uranium liberates $5.8\times 10^{8}\,J$ . Mass flow rate is computed herein:

$\dot m_{U}= \frac{2.758\times 10^{9}\,W}{5.8\times 10^{8}\,\frac{J}{g}}$

$\dot m_{U} = 4.755\,\frac{g}{s}$

The needed quantity of fuel per second is:

$\dot m_{fuel} = \frac{4.755\,\frac{g}{s} }{0.04}$

$\dot m_{fuel} = 118.875\,\frac{g}{s}$

The annual fuel consumption is now obtained:

$\Delta m_{fuel,year} = (0.119\,\frac{kg}{s})\cdot (\frac{3600\,s}{1\,h} )\cdot (\frac{24\,h}{1\,day} )\cdot(\frac{365\,days}{1\,year} )$

$\Delta m_{fuel,year} = 3,752,784\,\frac{kg}{year}$

The cost of fuel for the lifetime of the plant is:

$C_{lifetime} = (3,752,784\,\frac{kg}{year} )\cdot (\frac{0.453\,lb}{1\,kg} )\cdot [(20\,years)\cdot(\frac{12\,USD}{1\,lb} )+(20\,years)\cdot (\frac{65\,USD}{1\,lb} )$

$C_{lifetime} = 2,618,017,174\,USD$

6 0

3 years ago