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kirill115 [55]
4 years ago
14

As a photon from the Sun's core interacts with an atom or ion, the photon gives up its energy to the particle. But the excited p

article gives up this excess energy as another photon. Thus the core's energy is carried outward in a process called _____ transport.
Physics
1 answer:
Natalka [10]4 years ago
8 0

Answer:

Radiative transport

Explanation:

Radiative transportation takes place extensively in the core of the sun. It produces solar energy which are emitted outward from the core of the sun, and it comprises more than 70% of the sun's heat energy. At the core of the sun, the process of nuclear fusion takes where the hydrogen atoms combines with one another, and forms helium, and during this process, it releases a large amount of energy. The photons continuously interacts with the ions (or atoms) and releases this energy to the particles and the excited particles converts the excess amount of energy into another photon which are carried outward by the process of radiative transport.

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3 years ago
A person riding in a hot air air balloon basket ascends to 1000 ft above the ground. What happened to the atmospheric pressure a
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C) it decreased
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8 0
3 years ago
A 20-kg block slides down a fixed rough curved track The block has a speed of 5 0 m/s after its height above a horizontal surfac
Crank

Answer:

U = 102.8 J (100 J to two significant digits)

Explanation:

potential energy converted = 20(9.8)(1.8) = 352.8 J

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3 years ago
Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin.
Sergio039 [100]

Answer:

The distance is 1.69 m.

Explanation:

Given that,

First charge q_{1}= 3.8\times10^{-6}\ C

Second charge q_{2}=3.2\times10^{-6}\ C

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

E_{1}=E_{2}

\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}

\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}

\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}

x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}

Put the value into the formula

x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}

x=1.69\ m

Hence, The distance is 1.69 m.

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