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4 years ago

State how the sun transfers energy to earth

2 answers:

Nookie1986 [14]4 years ago

6 0

The sun transfers energy to Earth in the form of electromagnetic radiation.
That includes radio waves, heat, light, ultraviolet, and X-ray energy.

Mashutka [201]4 years ago

6 0

Radiation is the transfer of heat energy through space by electromagnetic radiation. Most of the electromagnetic radiation that comes to the Earth from the sun is in the form of visible light. Light is made of waves of different frequencies.

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During nuclear fission and fusion, matter that seems to disappear is actually converted into

iris [78.8K]

Answer:

Energy (I need one more brainlist can i has?)

Explanation:

- Nuclear fusion occurs when two light nuclei fuse together into a heavier nucleus

- Nuclear fission occurs when a heavy, unstable nucleus breaks apart into two or more lighter nuclei

In both processes, the mass of the products is always smaller than the mass of the initial nuclei. This means that part of the initial mass has been converted into something else: into energy, which is released in the process.

The amount of energy released in the process can be calculated by using the famous Einstein's equivalence:

where m is the difference between the mass of the product and the initial mass of the nuclei, and c is the speed of light.

8 0

4 years ago

Read 2 more answers

A spaceship is traveling at 24,000 m/sec. At T=5 sec, the rocket trusts are turned on. At T=55 sec, the spaceship reaches a spee

slava [35]

Answer:

Explanation:

Acceleration is the change in velocity of a body with respect to time;

Acceleration = change in velocity/change in time

change in velocity = 29,500 - 24,000

change in velocity= 5,500

Change in time = 55 - 5

change in time = 50secs

Substitute into the formula;

spaceships acceleration = 24000/50

spaceships acceleration = 480 m/s²

<em>Hence the spaceships acceleration is 480m/s²</em>

8 0

3 years ago

A 6.00 kg ball is dropped from a height of 12.0 m above one end of a uniform bar that pivots at its center. The bar has mass 5.0

Andre45 [30]

Answer:

The height of the other ball go after the collision is 2.304 m.

Explanation:

Given that,

Mass of ball = 6.00 kg

Height = 12.0 m

Mass of bar =5.00 kg

Length = 4.00 m

Suppose we need to calculate how high will the other ball go after the collision

We need to calculate the velocity of ball

Using formula of velocity

$v=\sqrt{2gh}$

$v=\sqrt{2\times9.8\times12.0}$

$v=15.33\ m/s$

We need to calculate the angular momentum

Using formula of angular momentum

$l_{before}=mvr$

Put the value into the formula

$l_{before}=6.00\times15.33\times2.0$

$l_{before}=183.96\ kgm^2/s$

We need to calculate the angular momentum

Using formula of angular momentum

$l_{after}=I_{t}\omega$

$l_{after}=(\dfrac{ml^2}{12}+m_{1}r^2+m_{2}r^2)\omega$

Put the value into the formula

$l_{after}=(\dfrac{5\times4.00^2}{12}+6.00\times2.0^2+6.00\times2.0^2)\omega$

$183.96=54.66\omega$

$\omega=\dfrac{183.96}{54.66}$

$\omega=3.36\ rad/sec$

After collision the ball leaves with velocity

We need to calculate the velocity after collision

Using formula of the velocity

$v= r\omega$

$v=2.0\times3.36$

$v=6.72\ m/s$

We need to calculate the height

Using formula of height

$h=\dfrac{v^2}{2g}$

Put the value into the formula

$h=\dfrac{(6.72)^2}{2\times9.8}$

$h=2.304\ m$

Hence, The height of the other ball go after the collision is 2.304 m.

5 0

3 years ago

. A student times a car traveling a distance of 2 m. She finds that it takes the car 5 s to

AveGali [126]

No, the car travels 1 metre in 5s at the start which is 0.2m/s, while the second meter it travels one metre in 8 seconds which is 0.125 m/s, the speed changes therefore it is not constant during the two metres the car travels

3 0

3 years ago

A 70 kg student jumps down to form a 1 m high platform. She forgets to bend her knees and her downward motion stops in 0.02 seco

34kurt

Answer:

15,505 N

Explanation:

Using the principle of conservation of energy, the potential energy loss of the student equals the kinetic energy gain of the student

-ΔU = ΔK

-(U₂ - U₁) = K₂ - K₁ where U₁ = initial potential energy = mgh , U₂ = final potential energy = 0, K₁ = initial kinetic energy = 0 and K₂ = final kinetic energy = 1/2mv²

-(0 - mgh) = 1/2mv² - 0

mgh = 1/2mv² where m = mass of student = 70kg, h = height of platform = 1 m, g = acceleration due to gravity = 9.8 m/s² and v = final velocity of student as he hits the ground.

mgh = 1/2mv²

gh = 1/2v²

v² = 2gh

v = √(2gh)

v = √(2 × 9.8 m/s² × 1 m)

v = √(19.6 m²/s²)

v = 4.43 m/s

Upon impact on the ground and stopping, impulse I = Ft = m(v' - v) where F = force, t = time = 0.02 s, m =mass of student = 70 kg, v = initial velocity on impact = 4.43 m/s and v'= final velocity at stopping = 0 m/s

So Ft = m(v' - v)

F = m(v' - v)/t

substituting the values of the variables, we have

F = 70 kg(0 m/s - 4.43 m/s)/0.02 s

= 70 kg(- 4.43 m/s)/0.02 s

= -310.1 kgm/s ÷ 0.02 s

= -15,505 N

So, the force transmitted to her bones is 15,505 N

3 0

3 years ago