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3 years ago

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

2 answers:

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]3 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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two circular plates, each with a radius of 8.22 cm, have equal and opposite charges of magnitude 3.052 μc. calculate the electri

PtichkaEL [24]

If the separation distance is doubled, then the electric field decreases by a factor of 4.

<h3>What is the electric field strength?</h3>

We know that the electric field strength is known to depend on the magnitude of the charge and the distance of separation. We know that the electric field refers to the region in which the influence of a charge is felt. Recall that a charge is a specie that is positively or negatively charged. The charge on a specie must always be shown by its sign.

We know that the electric field is the region in space where the influence of a charge can be felt. If a charge is placed in the vicinity of another charge, the second charge would experience a force due to the presence of the first charge. This is because the second charge was brought into the electric field of the first charge.

Thus we know that;

E = Kq/r^2

Where;

E = electric field strength

q = magnitude of charge

r = distance of separation

Now;

E = 9.0* 10^9 * 3.052 * 10^-6/(8.22 * 10^-2)^2

E = 4 N/C

Given that the electric filed strength is inversely proportional to the distance of separation, when the distance between the charges is doubled, the electric field decreases by a factor of 4.

Learn more about electric field strength:brainly.com/question/15170044?

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1 year ago

you want to get more physically fit but often make excuses for why you don't have time to be active what strategy can help with

Anna35 [415]

<span>Find an activity you enjoy and make it a priority in your schedule.</span>

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3 years ago

All stars go through a lifecycle.

Komok [63]

Mira is much bigger than the Sun.

Only very massive stars will go through a supernova stage, causing the outer layer to explode away and the core to collapse in on itself, becoming very dense.

5 0

2 years ago

Read 2 more answers

Assuming a vertical trajectory with no drag, derive the applicable form of the rocket equation for this application

VARVARA [1.3K]

Answer:

The vertical trajectory is governed by Ordinary Differential Equation.

Time derivatives of each state variables.

d(d)/dt = v, d(m)/dt = -d(m-fuel)/dt, d(v)/dt = F/m.

Where V is velocity positive upwards, t is time, m is mass, m-fuel is fuel mass, F is Total force, positive upwards.

Therefore,

F = -mg - D + T, If V is positive and

F = -mg + D - T, If T is negative.

D is drag and the questions gave it as zero.

Explanation:

The two sign cases in derivative equations above are required because F is defined positive up, so the drag D and thrust T can subtract or add to F depending in the sign of V . In contrast, the gravity force contribution mg is always negative. In general, F will be some function of time, and may also depend on the characteristics of the particular rocket. For example, the T component of F will become zero after all the fuel is expended, after which point the rocket will be ballistic, with only the gravity force and the aerodynamic drag force being p

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3 years ago

Anna applies a force of 19.5 newtons to push a book placed on a table. If the normal force of the book is 51.7 newtons, what is

andrey2020 [161]

Considering that the book is moving with constant speed, the force applied by Anna must be the same that the friction force:

$F = F_R = F\cdot \mu_k\cdot N$

If we clear the previous equation:

$\mu_k = \frac{F}{F_R} = \frac{19.5\ N}{51.7\ N} = \bf 0.38$

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3 years ago

Read 2 more answers