All categories

4 years ago

Explain how energy is conserved when nuclear fission or fusion occurs

Physics

2 answers:

andriy [413]4 years ago

6 0

Answer:

The energy is conserved in the nuclear fission when the nuclear reactor mass releases the kinetic energy in the reaction and then source is known as the nuclear binding energy.

By using the Einstein mass energy equation that is E = m×c², we can easily determine the actual amount of the released energy. The nuclear fission is the process of subdividing the heavy mass of the atomic nucleus like uranium.

The nuclear fusion basically violate the law (Conservation of mass). During the nuclear fusion process the matter are not conserve because when the nuclei gets fuse then its mass are converted into the energy and then it is released.

ollegr [7]4 years ago

5 0

The mass lost in the nuclear reaction is all converted to energy.

You might be interested in

The river narrows at a rapids from a width of 12 m to a width of only 5.8 m. The depth of the river before the rapids is 2.7 m;

Alisiya [41]

Answer:

7.89 m/s

Explanation:

Given that

Width of the river, b1 = 12 m

Width of the river, b2 = 5.8 m

Depth of the river, d1 = 2.7 m

Depth of the river, d2 = 0.85 m

Speed of the river, v1 = 1.2 m/s

Speed of the river, v2 = ?

Area of the river before the rapid, a1 = 12 * 2.7 = 32.4 m²

Area of the river after the rapid, a2 = 5.8 * 0.85 = 4.93 m²

To solve this question, we use a relation between the speed of the river and the volume of the river. We say,

Area1 * velocity1 = Area2 * velocity2, and when we substitute the values for each other we have

32.4 * 1.2 = 4.93 * v2

38.88 = 4.93v2

v2 = 38.88 / 4.93

v2 = 7.89 m/s

Therefore, the speed of the river after the rapid is 7.89 m/s

6 0

3 years ago

An object with mass 3.5 kg is attached to a spring with spring stiffness constant k = 270 N/m and is executing simple harmonic m

Elanso [62]

Answer:

Part a)

A = 0.066 m

Part b)

maximum speed = 0.58 m/s

Explanation:

As we know that angular frequency of spring block system is given as

$\omega = \sqrt{\frac{k}{m}}$

here we know

m = 3.5 kg

k = 270 N/m

now we have

$\omega = \sqrt{\frac{270}{3.5}}$

$\omega = 8.78 rad/s$

Part a)

Speed of SHM at distance x = 0.020 m from its equilibrium position is given as

$v = \omega \sqrt{A^2 - x^2}$

$0.55 = 8.78 \sqrt{A^2 - 0.020^2}$

$A = 0.066 m$

Part b)

Maximum speed of SHM at its mean position is given as

$v_{max} = A\omega$

$v_{max} = 0.066(8.78) = 0.58 m/s$

4 0

3 years ago

(11%) Problem 5: A submarine is stranded on the bottom of the ocean with its hatch 25 m below the surface. In this problem, assu

V125BC [204]

Answer:

F = 1.24*10^4 N

Explanation:

Given

Depth of the ship, h = 25 m

Density of water, ρ = 1.03*10^3 kg/m³

Diameter of the hatch, d = 0.25 m

Pressure of air, P(air) = 1 atm

Pressure of water =

P(w) = ρgh

P(w) = 1.03*10^3 * 9.8 * 25

P(w) = 2.52*10^5 N/m²

P(net) = P(w) + P(air) - P(air)

P(net) = P(w)

P(net) = 2.52*10^5 N/m²

Remember,

Pressure = Force / Area, so

Force = Area * Pressure

Area = πr² = πd²/4

Area = 3.142 * 0.25²/4

Area = 3.142 * 0.015625

Area = 0.0491 m²

Force = 0.0491 * 2.52*10^5

F = 12373 N

F = 1.24*10^4 N

5 0

3 years ago

Read 2 more answers

Un globo de aire caliente tiene un volumen de 560 mL a la presión atmosférica normal y una temperatura del aire de 400 ºC. Cuand

dalvyx [7]

Explanation:

charles law V1/T1 =V2/T2

560 x 673 =V2/973

376880 = V2/973

V2 = 376880 x 973 = 366704240mL

4 0

3 years ago

If a 80kg diver jumps off of a 5 m high dive into a regulation diving pool, how much should the temperature of the pool go up?

Agata [3.3K]

Answer:

The answer cannot be determined.

Explanation:

The energy of the diver when he hits the pool will be equal to its potential energy $mgh$ , and for the temperature of the pool to rise up, this energy has to be converted into the heat energy of the pool.

The change in temperature ${\Delta}T$ then will be

${\Delta}T=\frac{{\Delta}Q}{mc} .$

Where m is the mass of water in the pool, c is the specific heat capacity of water, and ${\Delta}Q$ is the added heat which in this case is the energy of the diver.

Since we do not know the mass of the water in the pool, we cannot make this calculation.

7 0

3 years ago