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Kitty [74]
4 years ago
10

In the nuclear fission process mass is converted into energy. Determine the total mass that must be converted to energy in one y

ear by a 1 Giga-Watt nuclear if we a. assume 100% efficiency. b, assume only 30 % efficiency (nuclear energy to electrical energy conversion).
Physics
1 answer:
brilliants [131]4 years ago
8 0

Answer:

1) Mass that needs to be converted at 100% efficiency is 0.3504 kg

2) Mass that needs to be converted at 30% efficiency is 1.168 kg

Explanation:

By the principle of mass energy equivalence we have

E=mc^{2}

where,

'E' is the energy produced

'm' is the mass consumed

'c' is the velocity of light in free space

Now the energy produced by the reactor in 1 year equals

Energy=Power\times time\\\\\therefore Energy=1\times 10^{9}\times 365\times 24\times 3600\\\\Energy=31.536\times 10^{15}Joules

Thus the mass that is covertred at 100% efficiency is

mass=\frac{Energy}{c^{2}}\\\\mass=\frac{31.536\times 10^{15}}{(3\times 10^{8})^{2}}\\\\mass=\frac{31.536\times 10^{15}}{9\times 10^{16}}\\\\\therefore mass=0.3504kg

Part 2)

At 30% efficiency the mass converted equals

mass|_{30}=\frac{mass|_{100}}{0.3}\\\\mass|_{30}=\frac{0.3504}{0.3}\\\\mass|_{30}=1.168kg

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6 0
3 years ago
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If a person weighs 717 N on earth and 5320 N on the surface due to gravity on that planet? of a nearby planet, what is the accel
Studentka2010 [4]

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4 0
4 years ago
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What force is described as the attraction between a sample of matter and all other matter in the universe?
Vinil7 [7]

Answer:

Gravitational Force

Explanation:

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Newton's Law of Universal Gravitation states the force of attraction between two masses m₁ and m₂ in the universe is directly proportional to the product of their masses and inversely proportional to the square of their distance apart.

                                 F = G\frac{m_{1}m_{2}}{r^2}

Where;

F is the gravitational force,

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m1 and m2 are the masses of the objects,

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4 0
4 years ago
Two trains leave the station at the same time, one heading east and the other west. the eastbound train travels at 85 miles per
kakasveta [241]
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3 0
3 years ago
A raft of mass 199 kg carries two swimmers of mass 52 kg and 70 kg. The raft is initially floating at rest. The two swimmers sim
Minchanka [31]

To solve this problem we will apply the concept related to the conservation of the Momentum. We will then start considering that the amount of initial momentum must be equal to the amount of final momentum. Considering that all the objects at the initial moment have the same initial velocity (Zero, since they start from rest) the final moment will be equivalent to the multiplication of the mass of each object by the velocity of each object, so

Initial Momentum = Final Momentum

(m_B+m_1+m_2)v_i = m_1v_1+m_2v_2+m_Bv_B

Here,

m_B =  mass of Raft

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m_2 = Mass of swimmers 2

v_i = Initial velocity (of the three objects)

v_B = Velocity of Raft

Replacing,

(199+52+70)*0 = (52)(4)+(70)(-4)+199v_B

Solving for v_B

vB = \frac{72}{199}

v_B = 0.3618m/s

Therefore the velocity the rarft start to move is 0.3618m/s

5 0
4 years ago
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