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Ray Of Light [21]
4 years ago
5

Energy released by fusion in the sun is initially in the form of

Physics
2 answers:
bogdanovich [222]4 years ago
4 0
Energy released by fusion in the sun is initially in the form of gamma rays.

Gamma rays arise from the radioactive decay of nuclei. They are penetrating electromagnetic radiations consisting of very high energy photons.
Gamma rays are ionizing radiations and have very serious biological dangers and hazards (due to their ability of ionizing the atoms).
iVinArrow [24]4 years ago
4 0

Answer:

Initially energy is in the form of Radiations

Explanation:

Sun is made up of gases in which major proportion is hydrogen gas.

Here we know that surface of sun is vert hot and its temperature is very large

So due to this high temperature hydrogen gas combined and form helium nuclei, this change in the nuclei reaction is known as nuclear reaction.

In this nuclear reaction we can see that two light nuclei will combine and form a big stable nuclei. This type of reaction is known as nuclear fusion and energy released in this type of nuclear fusion is always in form of gamma radiations.

Now these gamma radiations are absorbed by other charged particles and those charged particles are then accelerated to higher energy.

So initial form of this energy is Gamma Radiations

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How can the IMA of a first- class lever be increased?
Dimas [21]
IMA = Ideal Mechanical Advantage

First class lever = > F1 * x2 = F2 * x1

Where F1 is the force applied to beat F2. The distance from F1 and the pivot is x1 and the distance from F2 and the pivot is x2

=> F1/F2 = x1 /x2

IMA = F1/F2 = x1/x2

Now you can see the effects of changing F1, F2, x1 and x2.

If you decrease the lengt X1 between the applied effort (F1) and the pivot,  IMA decreases.

If you increase the length X1 between the applied effort (F1) and the pivot, IMA increases.

If you decrease the applied effort (F1) and increase the distance between it and the pivot (X1) the new IMA may incrase or decrase depending on the ratio of the changes.

If you decrease the applied effort (F1) and decrease the distance between it and the pivot  (X1) IMA will decrease.

Answer: Increase the length between the applied effort and the pivot.
4 0
4 years ago
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Find the work performed when the given force f f is applied to an object, whose resulting motion is represented by the displacem
Ahat [919]

work = force x distance

fd foot pounds

3 0
3 years ago
3. Waxing means "growing." Waning means
Sonja [21]

Answer:

I think it is the last one.

Explanation:

I am not sure because i am stuck on this one, too.

4 0
3 years ago
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. An object whose mass is 375 lb falls freely under the influence of gravity from an initial elevation of 253 ft above the surfa
lozanna [386]

Answer:

(a) Vf = 128 ft/s

(b) K.E = 122.8 Btu

Explanation:

(a)

In order to find the velocity of the object just before striking the surface of earth or the final velocity, we use 3rd equation of motion:

2gh = Vf² - Vi²

where,

g = 32.2 ft/s²

h = height = 253 ft

Vf = Final Velocity = ?

Vi = Initial Velocity = 10 ft/s

Therefore,

(2)(32.2 ft/s²)(253 ft) = Vf² - (10 ft/s)²

16293.2 ft²/s² + 100 ft²/s² = Vf²

Vf = √(16393.2 ft²/s²)

<u>Vf = 128 ft/s</u>

<u></u>

(b)

The kinetic energy of the object before it hits the surface of earth is given by:

K.E = (0.5)(m)(Vf)²

where,

m = mass of object = 375 lb

K.E = Kinetic energy of object before it strikes the surface of earth = ?

Therefore,

K.E = (0.5)(375 lb)(128 ft/s)²

K.E = 3073725 lb.ft²/s²

Now, converting this to Btu:

K.E = (3073725 lb.ft²/s²)(1 Btu/25037 lb.ft²/s²)

<u>K.E = 122.8 Btu</u>

3 0
4 years ago
A toy car moves 8 min 4 s at the constant velocity. What is the car's velocity?
Schach [20]

Explanation:

<u>Formula:</u>

velocity = (d \div t)

<u>d = distance given</u>

<u>t</u><u> </u><u>=</u><u> </u><u>the amount of time </u><u>given</u>

<u>Substitute the given values into the formula for velocity</u><u>:</u>

v = 8 \div 4

velocity is shortened for v.

8 (distance) divided by 4 (time) equals the velocity.

<u>Solve:</u>

2 = 8 \div 4

The velocity of the toy car equals: B. 2 m/s.

6 0
2 years ago
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