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3 years ago

Explain nuclear fusion briefly

Physics

1 answer:

nikdorinn [45]3 years ago

6 0

Answer:

Nuclear fusion is the process in which two atomic nuclei are fused together into a larger single nucleus, releasing energy in the process. ... In proton-proton fusion, four hydrogen atoms are fused into a single helium-4 atom, releasing a lot of energy in the process.

Explanation:

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The inner cylinder of a long, cylindrical capacitor has radius r and linear charge density +λ. It is surrounded by a coaxial cyl

Ulleksa [173]

Hi there!

We can begin by using the equation for energy density.

$U = \frac{1}{2}\epsilon_0 E^2$

U = Energy (J)

ε₀ = permittivity of free space

E = electric field (V/m)

First, derive the equation for the electric field using Gauss's Law:
$\Phi _E = \oint E \cdot dA = \frac{Q_{encl}}{\epsilon_0}$

Creating a Gaussian surface being the lateral surface area of a cylinder:
$A = 2\pi rL\\\\E \cdot 2\pi rL = \frac{Q_{encl}}{\epsilon_0}\\\\Q = \lambda L\\\\E \cdot 2\pi rL = \frac{\lambda L}{\epsilon_0}\\\\E = \frac{\lambda }{2\pi r \epsilon_0}$

Now, we can calculate the energy density using the equation:
$U = \frac{1}{2} \epsilon_0 E^2$

Plug in the expression for the electric field and solve.

$U = \frac{1}{2}\epsilon_0 (\frac{\lambda}{2\pi r \epsilon_0})^2\\\\U = \frac{\lambda^2}{8\pi^2r^2\epsilon_0}$

Now, we can integrate over the volume with respect to the radius.

Recall:
$V = \pi r^2L \\\\dV = 2\pi rLdr$

Now, we can take the integral of the above expression. Let:
$r_i$ = inner cylinder radius

$r_o$ = outer cylindrical shell inner radius

Total energy-field energy:

$U = \int\limits^{r_o}_{r_i} {U_D} \, dV = \int\limits^{r_o}_{r_i} {2\pi rL *U_D} \, dr$

Plug in the equation for the electric field energy density and solve.

$U = \int\limits^{r_o}_{r_i} {2\pi rL *\frac{\lambda^2}{8\pi^2r^2\epsilon_0}} \, dr\\\\U = \int\limits^{r_o}_{r_i} { L *\frac{\lambda^2}{4\pi r\epsilon_0}} \, dr\\$

Bring constants in front and integrate. Recall the following integration rule:
$\int {\frac{1}{x}} \, dx = ln(x) + C$

Now, we can solve!

$U = \frac{\lambda^2 L}{4\pi \epsilon_0}\int\limits^{r_o}_{r_i} { \frac{1}{r}} \, dr\\\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(r)\left \| {{r_o} \atop {r_i}} \right. \\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} (ln(r_o) - ln(r_i))\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})$

To find the total electric field energy per unit length, we can simply divide by the length, 'L'.

$\frac{U}{L} = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})\frac{1}{L} \\\\\frac{U}{L} = \boxed{\frac{\lambda^2 }{4\pi \epsilon_0} ln(\frac{r_o}{r_i})}$

And here's our equation!

3 0

2 years ago

What is the velocity of the buggy? <br><br> At 20 seconds the buggy will have a position of ____ ?

lbvjy [14]

At 20 seconds it will be 12.6 because at 10 seconds it was as at approximately 6.3 so we times it by 2 to get the 20s

6 0

3 years ago

Why should galaxy collisions have been more common in the past than they are today?

Roman55 [17]

Answer:

I think d is the answer haha

7 0

4 years ago

Name on contact force and one field force acting on you right now

kati45 [8]

Answer:

Contact forces are forces that require the actual contact (touching) of two pieces of matter. There are a variety of contact forces. A very common one is friction. Anytime that two surfaces are in contact with one another, there is friction between the two surfaces. A field force is a force that works at a distance. No touching is required. Gravity is a good example of a field force, because it works whether or not an object is touching something or touching nothing at all.

8 0

3 years ago

H e l p <br> P l e a s e.

gtnhenbr [62]

Answer:

I still cant see no matter how much I zoom in!???!

Explanation:

To Small

8 0

3 years ago

Read 2 more answers