9.00 V is applied to a wire with a

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

maksim [4K]

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.

6 0

3 years ago

Read 2 more answers

The pressure on a volume of liquid V = 1.0 mº at the surface is approximately equal to the atmospheric pressure Patm = 1.00 x 10

Alexus [3.1K]

Answer:-2.86*10⁻⁴

Explanation: Use the equation change in volume = (change in pressure * original volume) / Bulks Modulus. ΔV = (-Δp*V₀) / B

Plugging in your numbers, you should get ΔV = (-2.29*10⁷*1) / (8*10¹⁰) = -2.86*10⁻⁴

ΔP = P₂-P₁ ----> ΔP = 2.30*10⁷ - 1.00*10⁵ = 2.29*10⁷

3 0

3 years ago

Jake is helping Fin push a box at a constant velocity up an incline that makes an angle of 30.0° above the horizontal by applyin

andre [41]

Given data

The angle of inclination of the plane is theta = 30 degree

The applied force in the inclined plane is F = 94 N

The distance moved in the inclined plane is d = 2.30 m

The coefficient of kinetic friction is u_k = 0.280

The free-body diagram of the above configuration is shown below:

Here, the normal reaction force on the box is N, the acceleration due to gravity is denoted as g, the friction force on the box is F_f, and the mass of the box is denoted as m.

(a)

The expression for the work done by the pushing force is given as:

$W=Fd$

Substitute the value in the above equation.

$\begin{gathered} W=94\text{ N}\times2.30\text{ m} \\ W=216.2\text{ J} \end{gathered}$

Thus, the work done by the pushing force is 216.2 J.

(b)

The box is moving at the constant velocity, therefore, the pushing force will be equal to the frictional force and the component of the gravitational force in the inclined plane.