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Amiraneli [1.4K]
3 years ago
15

A certain nuclear power plant is capable of producing 1.2×10^9 W of electric power. During operation of the reactor, mass is con

verted to energy. How much mass is converted per hour if the efficiency of the plant is 30 percent? (Answer is in kilograms)
Physics
1 answer:
alukav5142 [94]3 years ago
6 0

Answer:

0.00016 kg

Explanation:  

Given:

Power = P = 1.2 × 10⁹ Watts

Power =  work done / Time

efficiency = 0.30

Input power = 1.2 × 10⁹ / 0.30 =  4  × 10⁹ W

Energy =  4  × 10⁹ x 60 x 60 = 1.44 x 10¹³ joules

E = m c² , where c is the speed of light and m is the mass.

⇒ mass = m = E / c²  = (1.44 x 10¹³) / (3 × 10⁸ )²

                                   = 0.00016 kg

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Rest and motion are relative terms why?​
krok68 [10]

Answer:

Rest and motion are the relative terms because they depend on the observer's frame of reference. So if two different observers are not at rest with respect to each other, then they too get different results when they observe the motion or rest of a body.

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The light of that star actually gives off
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a brick is suspended above the ground at a height of 6.6 m. it has a mass of 5.3 kg. what is the potential energy of the brick
Svetradugi [14.3K]
The formula for potential energy is
E(p) = mgh

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Therefore energy = (5.3)(9.8)(6.6)
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8 0
3 years ago
2 (a) A plane from airport A flies 280 km to the east to airport B. The plane then travelled north to airport C, 190 km away. (i
Mekhanik [1.2K]

Answer:

See below

Explanation:

See attached diagram

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4 0
1 year ago
Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge
NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is V=\frac{1}{4\pi\epsilon} \frac{q}{d}

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as-E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero  between the two charges,but we can find the points situated on the axial  line but  outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining  two charges  where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

5 0
3 years ago
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