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3 years ago

How is sun related to nuclear, electromagnetic, and heat energy?

2 answers:

7 0

The nuclear energy is the source of energy production on the sun, then this travels to the Earth in the form of Electromagnetic radiation and as it heats up the other bodies in solar system, it gets converted into the heat energy.

Explanation:

The sun is considered to be the ultimate source of energy for the whole solar system. The energy produced in the sun reaches all the other planets in the solar system. The sunlight is a very big source of energy for the Earth also.

The energy produced on the sun is due to the nuclear reaction due to the immense pressure and temperature. This nuclear energy from the sun travels to the other bodies in the solar system by means of the electromagnetic radiations.

The electromagnetic radiations from the sun carry the heat energy to the bodies and as they reach the target, the target is heated up due to the conversion of the energy from the sun into the heat energy.

Thus, The nuclear energy is the source of energy production on the sun, then this travels to the Earth in the form of Electromagnetic radiation and as it heats up the other bodies in solar system, it gets converted into the heat energy.

Learn More:

1. Acceleration of the box under friction brainly.com/question/7031524

2. The kinetic energy of an object depends on the brainly.com/question/137098

3. The approximate time taken to walk for 2000 miles brainly.com/question/3785992

Answer Details:

Grade: High School

Subject: Physics

Chapter: Energy

Keywords:

sun, energy, nuclear, electromagnetic, heat, nuclear reaction, immense, pressure, temperature, source, solar system, planets.

N76 [4]3 years ago

3 0

<h3>Answer;</h3>

The sun creates energy through nuclear fusion. Kinetic energy creates heat and light.

<h3>Explanation;</h3>

Solar energy from the sun is the primary source of energy on the surface of life that is important for sustaining life on earth.
Nuclear fusion occurring in the sun involves a nuclear reaction in which hydrogen nuclei fuse together to form helium nuclei and release energy.
Electromagnetic energy from the sun comes to Earth in the form of radiation. This energy travels through space at a speed equivalent to that of other electromagnetic radiations that is the speed of light.
The electromagnetic energy from nuclear fusion in the sun generates kinetic energy that in turn creates light and heat.

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A car with a mass of 1,500 kg is traveling at a speed of 30 m/s. What force must be applied to stop the car in 3 seconds?

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Explanation:

F = ma

F = m Δv / Δt

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F = -15,000 N

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3 years ago

A potential energy function is given by u(x)=(3.00n)xâ(1.00n/m2)x3. at what position or positions is the force equal to zero?

qwelly [4]

I believe the correct form of the energy function is:

u (x) = (3.00 N) x + (1.00 N / m^2) x^3

or in simpler terms without the units:

u (x) = 3 x + x^3

Since the highest degree is power of 3, therefore there are two roots or solutions of the equation.

Since we are to find for the positions x in which the force equal to zero, u (x) = 0, therefore:

3 x + x^3 = u (x)

3 x + x^3 = 0

Taking out x:

x (3 + x^2) = 0

So one of the factors is x = 0.

Finding for the other two factors, we divide the two sides by x and giving us:

x^2 + 3 = 0

x^2 = - 3

x = sqrt (- 3)

x = - 1.732 i, 1.732 i

The other two roots are imaginary therefore the force is only equal to zero when the position is also zero.

Answer:

x = 0

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3 years ago

The 3.00 kg cube in fig. 15-47 has edge lengths d 6.00 cm and is mounted on an axle through its center. a spring (k 1200 n/m con

ira [324]

The Period of the resulting shm will be T=39.7

Explanation:

Given data

m=3kg

d=.06m

k=1200 N/m

Θ=3 °

T=?

we have the formulas,

I = (1/6)Md2

F = ma

F = -kx = -(mω2x)

k = mω2 τ = -d(FgsinΘ)

T=2 x 3.14/ √(m/k)

Solution for the given problem would be,

F=-Kx (where x= dsin Θ)

F=-k dsin Θ

F=-(1200)(.06)sin(3 °)

F=-10.16N

By newton's second law.

F = ma

a= F/m

a=(-10.16N)/3

a=3.38

using the k=mω value

k=mω

ω=k/m

ω=1200/3

ω=400

Using F = -kx value

x = F/-k

x=(-10.16)/1200

x=0.00847m

Restoring the torque value

τ = -dmgsinΘ where( τ = Iα so.).. Iα = -dmgsinΘ α = -(.06)(4)α =

α =(.06)(4)(9.81)sin(4°)

α=-1.781

Rotational to linear form

a = αr

r = .1131 m

a=-1.781 x .1131 m

a=-0.2015233664

Time Period

T=2 x 3.14/ √(m/k)

T=6.28/√(3/1200)

T=6.28/0.158

T=39.7

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3 years ago