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Each fission of a U235 nucleus produces 200 Mev which in terms of Joules is 3.2 x 10-11 Joules. This is a very small amount, which shows just how many fissions are occurring every second, for a reactor which produces 3000 Mw thermal
60 grains = 0.003888 kg 3000 feet per second = 914.4 m/s Kinetic Energy = 0.5 x Mass x Velocity^2 KE = 0.5 x 0.003888 x 914.4^2 = 1625.43158784 Joules
3000 joules(J)
Given the wavelength of the photons from above, 3000 nm you just calculate how many joules each photon has and divide that into 100 joules per second.
If this is about the mass-equivalent of energy, you need to do the following: convert the 10 kg to the equivalent energy in joules, using the formula e = m times c squared; then divide the energy you get by 1000 joules, to find how many seconds it will take you. To convert that to hours, divide the result by 3600 (since there are 3600 seconds in an hour). The "physics" answer above is not the biological answer because (wild guess) a pound of fat comes from 3000 calories. There is a lot of extra matter needed to support the energy in a person.
Each fission of a U235 nucleus produces 200 Mev which in terms of Joules is 3.2 x 10-11 Joules. This is a very small amount, which shows just how many fissions are occurring every second, for a reactor which produces 3000 Mw thermal
I will use this formula. Some conversion will be required. ( I only know specific heat iron in J/gC ) q(Joules) = mass * specific heat * change in temperature Celsius 3 kilograms cast iron = 3000 grams q = (3000 g)(0.46 J/gC)(120 C - 30 C) = 124200 Joules (1 kilojoule/1000 joules) = 124.2 kilojoules of energy needed ===========================
60 grains = 0.003888 kg 3000 feet per second = 914.4 m/s Kinetic Energy = 0.5 x Mass x Velocity^2 KE = 0.5 x 0.003888 x 914.4^2 = 1625.43158784 Joules
The solar flux is 120,000 terrawatts. Multiply that by the number of seconds in a year to get the total number of joules of energy hitting the Earth each year. It's a big number. A very very big number!
The voltage is not relevant, only the power (wattage). At 3 kW = 3000 watts, this means 3000 joules per second (J/s), as 1 watt = 1 J/s.CommentThe voltage is relevant! A 3kW electric fire, rated at 230 V, will only produce 3 kW at that rated voltage.
3000 joules(J)
3000 joules(J)
3000 joules(J)
The Dell Dimension 3000 made its debut in 2004.
1 BTU = 1.055 kilojoules. For a nuclear plant with an electrical output of say 1000 MWe, the reactor thermal output will be about 3000 MW (at 33 percent efficiency), or 3000 Mega joules/second, which is 3000 x 1000 kilojoules/sec, or 3000/1.055 x 1000 BTU/sec. this reduces to 2.84 x 106 BTU/second, Scale it according to the actual electrical output of the plant.
The specific heat of iron is either 0.46 or 0.45 Jolules/grams*Celsius, so......... q(in Joules) = mass * specific heat * Temp final - Temp. initial q = (65 grams Fe)(0.46 J/gC)(95 C - 25 C) = 2093 joules of energy
Trick question--The answer is none. Indeed, energy in the form of heat is produced by the car stopping. But no energy is required to stop it. The brake pads, and drums or discs absorb energy and get hot.