All categories

2 years ago

How does enormous energy get released from the sun?

Physics

1 answer:

Nezavi [6.7K]2 years ago

7 0

Enormous energy get released from the sun due to fusion reaction.

<h3>How does enormous energy get released from the sun?</h3>

Sun can generate enormous energy due to nuclear fusion. In this process, high temperature and pressure in the core of the sun force the nuclei to separate from their electrons which results in the formation of one Helium atom releasing a large amount of energy.

So we can conclude that enormous energy get released from the sun due to fusion reaction.

Learn more about energy here: brainly.com/question/13881533

#SPJ1

You might be interested in

What are the answers please help

aleksandr82 [10.1K]

Answer:

a.) The main scale reading is 10.2cm

b.) Division 7 = 0.07

c.) 10.27 cm

d.) 10.31 cm

e.) 10.24 cm

Explanation:

The figure depicts a vernier caliper readings

a.) The main scale reading is 10.2 cm

The reading before the vernier scale

b.) Division 7 = 0.07

the point where the main scale and vernier scale meet

c.) The observed readings is

10.2 + 0.07 = 10.27 cm

d.) If the instrument has a positive zero error of 4 division

correct reading = 10.27 + 0.04 = 10.31cm

e.) If the instrument has a negative zero error of 3 division

correct reading = 10.27 - 0.03 = 10.24cm

3 0

4 years ago

If your prediction turns out to be false your experiment has been a failure

nikdorinn [45]

This is false. Your hypothesis, or prediction, is just that: a prediction. Saying its a failure will result in bias.

3 0

4 years ago

Read 2 more answers

A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the lens strength (a.k.a, lens p

Kipish [7]

Answer:

20.0 cm

Explanation:

Here is the complete question

The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?

Solution

Now, the power of a lens, P = 1/f = 1/u + 1/v where f = focal length of lens, u = object distance from eye lens and v = image distance from eye lens.

Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.

Also, since the object is seen clearly, the distance from the eye lens to the retina equals the distance between the image and the eye lens. So, v = 2.00 cm = 0.02 m

Now, P' = 1/u + 1/v

1/u = P'- 1/v

1/u = 55.0 D - 1/0.02 m

1/u = 55.0 m⁻¹ - 1/0.02 m

1/u = 55.0 m⁻¹ - 50.0 m⁻¹

1/u = 5.0 m⁻¹

u = 1/5.0 m⁻¹

u = 0.2 m

u = 20 cm

So, at 55.0 dioptres, the closet object she can see is 20 cm from her eye.

8 0

3 years ago

A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the

lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

$l^2 = x^2 + y^2-----(1)$