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3 years ago

PLEASE HELP!!!!!!!

1 answer:

swat323 years ago

6 0

Nuclear fusion is the process by which multiple nuclei join together to form a heavier nucleus.

Nuclear fusion of light elements releases the energy that causes stars to shine and hydrogen bombs to explode.

Nuclear fusion of heavy elements (absorbing energy) occurs in the extremely high-energy conditions of supernova explosions.

Nuclear fusion in stars and supernovae is the primary process by which new natural elements are created.

It is this reaction that is harnessed in fusion power.

It takes considerable energy to force nuclei to fuse, even those of the lightest element, hydrogen.

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The moon has a mass of 7.35x1022 kg and is a lot farther away than is shown in textbooks. The mass of the

anyanavicka [17]

Answer:

$F=1.98\times 10^{20}\ N$

Explanation:

Given that,

The mass of a Moon, $M_m=7.35\times 10^{22}\ kg$

The mass of the Earth, $M_e=5.98\times 10^{24}\ kg$

The moon's mean orbit distance around the earth is, $r=3.84\times 10^8\ m$

We need to find the gravitational force exerted on the moon by the Earth.

The formula of gravitational force is given by :

$F=G\dfrac{M_mM_e}{r^2}\\\\F=6.67\times 10^{-11}\times \dfrac{7.35\times 10^{22}\times 5.98\times 10^{24}}{(3.84\times 10^8)^2}\\\\F=1.98\times 10^{20}\ N$

So, the required force is $1.98\times 10^{20}\ N$ .

3 0

3 years ago

An egg is dropped from a building that is 61 m high.

Allisa [31]

Answer:

Initial Velocity = 0 m/s

Final Velocity = 34.6 m/s

time = 3.5 s

Explanation:

The initial velocity must be zero since, the egg must be at rest initially, before dropping.

<u>Initial Velocity = 0 m/s</u>

Now, for time we use 2nd equation of motion:

h = Vi t + (1/2)gt²

where,

h = Height = 61 m

Vi = Initial Velocity = 0 m/s

g = 9.8 m/s²

t =time = ?

Therefore,

61 m = (0 m/s)(t) + (1/2)(9.8 m/s²)t²

t² = (61 m)(2)/(9.8 m/s²)

t = √(12.45 s²)

Now, for final velocity we will use 1st equation of motion:

Vf = Vi + gt

Vf = 0 m/s + (9.8 m/s²)(3.5 s)

Vf = 34.6 m/s

3 0

3 years ago

If a current of 2.4 a is flowing in a cylindrical wire of diameter 2.0 mm, what is the average current density in this wire?

Gnom [1K]

The average current density in the wire is given by:

$J=\frac{I}{A}$

where I is the current intensity and A is the cross-sectional area of the wire.

The cross-sectional area of the wire is given by:

$A=\pi r^2$

where r is the radius of the wire. In this problem, $r=\frac{d}{2}=\frac{2.0 mm}{2}=1.0 mm=0.001 m$ , so the cross-sectional area is

$A=\pi (0.001 m)^2=3.14 \cdot 10^{-6} m^2$

and the average current density is

$J=\frac{I}{A}=\frac{2.4 A}{3.14 \cdot 10^{-6} m^2}=7.64 \cdot 10^5 A/m^2$

8 0

3 years ago

Read 2 more answers

What happens when two forces act in the same direction?

Sholpan [36]

bigger acceleration......................

7 0

4 years ago

A body is traveling at 5.0 m/s along the positive direction of an x axis; no net force acts on the body. An internal explosion s

loris [4]

To solve this problem it is necessary to apply the concepts related to the conservation of momentum and conservation of kinetic energy.

By definition kinetic energy is defined as

$KE = \frac{1}{2} mv^2$

Where,

m = Mass

v = Velocity

On the other hand we have the conservation of the moment, which for this case would be defined as

$m*V_i = m_1V_1+m_2V_2$

Here,

m = Total mass (8Kg at this case)

$m_1=m_2 =$ Mass each part

$V_i =$ Initial velocity

$V_2 =$ Final velocity particle 2

$V_1 =$ Final velocity particle 1

The initial kinetic energy would be given by,

$KE_i=\frac{1}{2}mv^2$

$KE_i = \frac{1}{2}8*5^2$

$KE_i = 100J$

In the end the energy increased 100J, that is,

$KE_f = KE_i KE_{increased}$

$KE_f = 100+100 = 200J$

By conservation of the moment then,

$m*V_i = m_1V_1+m_2V_2$

Replacing we have,

$(8)*5 = 4*V_1+4*V_2$

$40 = 4(V_1+V_2)$

$V_1+V_2 = 10$

$V_2 = 10-V_1$ (1)

In the final point the cinematic energy of EACH particle would be given by

$KE_f = \frac{1}{2}mv^2$

$KE_f = \frac{1}{2}4*(V_1^2+V_2^2)$

$200J=\frac{1}{2}4*(V_1^2+V_2^2)$ (2)

So we have a system of 2x2 equations

$V_2 = 10-V_1$

$200J=\frac{1}{2}4*(V_1^2+V_2^2)$

Replacing (1) in (2) and solving we have to,

200J=\frac{1}{2}4*(V_1^2+(10-V_1)^2)

PART A: $V_1 = 10m/s$

Then replacing in (1) we have that

PART B: $V_2 = 0m/s$

8 0

3 years ago