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Usimov [2.4K]
3 years ago
5

Nuclear fusion has not yet been harnessed as an energy source because

Physics
2 answers:
Katen [24]3 years ago
7 0

Answer:

B

Explanation:

Energy is hard to control.

maksim [4K]3 years ago
6 0

Before getting an answer for it first we have to understand nuclear fusion.

Nuclear fusion is a thermo-nuclear reaction in which two  light unstable nuclei will form a heavy stable nuclei. In this process there will be some mass defect which will be converted into energy as per Einstein's mass energy equivalence theorem.

The theorem is stated as E = mc^{2} where c is the velocity of light and m is the mass converted into energy.

One take an example of fusion in sun where 4 hydrogen atoms combine to form a helium nucleus which are explained below-

                                                      H_{1}^{1}+H_{1} ^{1} =H_{1} ^{2} +e_{+1} ^{0} +energy

                                                      H_{1} ^{2} +H_{1} ^{1} =He_{2} ^{3} +energy

                                                      He_{2} ^{3}+H_{1} ^{1} =He_{2} ^{4} +e_{+1} ^{0} +energy

                                                       -----------------------------------------------------------------------

                                                        4H_{1} ^{1} = He_{2} ^{4} +2e_{+1} ^{0} +energy

Here e_{+1} ^{0} is the positron.

In this process very high temperature is needed which is approximately equal to the temperature of the sun i.e 10^{6} K

Such temperature is very difficult to initiate the reaction on the earth surface. It should be carried out in an sustainable way also .Otherwise It will cause nuclear hazards.

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A 1.0-kilogram object moving east with a velocity of 10 meters per second collides with a 0.50-kilogram object that is at rest.
NeTakaya

Answer:

10kg m/s

Explanation:

We can use the conversation of momentum for this question. Essentially the momentum before and after the crash will remain the same. We can use the formula P = mv to solve.

P = 1 * 10

P = 10kg m/s

Best of Luck!

3 0
3 years ago
I tried 1760km but it says it’s wrong
Lemur [1.5K]

Answer:

hmmmm

Explanation:

3 0
3 years ago
Which of the following is not an example of projectile motion?
Ilia_Sergeevich [38]
What are the answers?
in order to get help i would need the answers :)
4 0
4 years ago
Read 2 more answers
In Ampere's law, ∮????⃗∙????????⃗=????0???? the direction of the integration around the path:
kupik [55]

Answer:

C) must be such as to follow the magnetic field lines.

Explanation:

Ampere's circuital law helps us to calculate magnetic field due to a current carrying conductor. Magnetic field due to a current forms closed loop around the current . If a  net current of value I creates a magnetic field B around it , the line integral of magnetic field around a closed path becomes equal to μ₀ times the net current . It is Ampere's circuital law . There may be more than one current passing through the area enclosed by closed curve . In that case we will take net current by adding or subtracting them according to their direction.

It is expressed as follows

∫ B.dl = μ₀ I . Here integration is carried over closed path . It may not be circular in shape. The limit of this integration must follow magnetic field lines.

the term ∫ B.dl is called line integral of magnetic field.

3 0
4 years ago
3.1 * Consider a gun of mass M (when unloaded) that fires a shell of mass m with muzzle speed v. (That is, the shell's speed rel
kramer

Answer:

v_s=\frac{v}{1+\frac{m}{M} }

Explanation:

Let the shells speed with respect to ground be v_s

Let the shells speed with respect to ground be v_g

mass of shell, m

mass of gun, M

The relative velocity of shell with respect to a free unconstrained gun, v

According to the Newton's third law of motion the direction of the velocity of gun and shell will be in the opposite direction.

So, the relation between the relative velocity and their individual velocity will be:

v=v_s+v_g ......................(1)

<u>And according to the conservation of momentum (as the condition is very close to the elastic collision):</u>

M.v_g-m.v_s=0

substitute the value of v_g from equation (1)

M\times (v-v_s)=m\times v_s

M.v-M.v_s=m.v_s

M.v=M.v_s+m.v_s

v_s=\frac{M.v}{(M+m)}

v_s=\frac{v}{(\frac{(M+m)}{M}) }

v_s=\frac{v}{1+\frac{m}{M} }

5 0
3 years ago
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