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4 years ago

Which elements are the main components of the Sun?

Physics

2 answers:

dybincka [34]4 years ago

8 0

The answer is hydrogen and helium

hope this helped (:

Ksenya-84 [330]4 years ago

4 0

<u>Answer:</u> The main components of the Sun are Helium and Hydrogen.

<u>Explanation:</u>

Sun is the only star present in Our Solar System around which 8 planets revolve.

Sun is made up of many elements, some of them are: Hydrogen, Helium, Carbon, Nitrogen and Oxygen. The percentages of these elements by mass in Sun are:

Hydrogen: 70%

Helium: 28%

Carbon, Nitrogen and Oxygen: 1.5%

Other elements: 0.5%

In Sun, nuclear fusion reaction takes place where hydrogen undergoes fusion with other hydrogen nuclei to form helium nuclei. Equations for nuclear fusion are given as:

$_1^1\textrm{H}+_1^1\textrm{H}\rightarrow _1^2\textrm{H}+_1^0e+\gamma\\\\_1^2\textrm{H}+_1^1\textrm{H}\rightarrow _2^3\textrm{He}-+\gamma\\\\_2^3\textrm{He}+_2^3\textrm{He}\rightarrow _2^4\textrm{He}+2_1^1\textrm{H}\\\\_1^0e+_{-1}^0e\rightarrow \gamma$

Hence, the main components of the Sun are Helium and Hydrogen.

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What current is needed to generate the magnetic field strength of 5.0×10−5T at a point 1.5 cm from a long, straight wire? Expres

mixer [17]

Answer:

3.7 A

Explanation:

Parameters given:

Magnetic field strength, B = 5 * 10^(-5) T

Distance of magnetic field from wire, r = 1.5 cm = 0.015 m

The magnetic field, B, due to a current, I, flowing a wire is given as:

B = (μ₀*I) / 2πr

Where μ₀ = permeability of free space

To get the current, I, we make I the subject of the formula:

I = (2πr * B) / μ₀

I = (2 * 3.142 * 5 * 10^(-5)) / (1.25663706 × 10^(-6))

I = 3.7 A

4 0

4 years ago

An electron is accelrated by a unifor electric field (1000v/m) pointing vertically upward. Use energy methods to get the magnitu

ExtremeBDS [4]

Explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence, $qE \times S + mg \times S = 0.5 \times mv^{2}$

$1.6 \times 10^{-19} \times 1000 + 9.1 \times 10^{-31} \times 9.8 \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2}$

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight and the above equation will be as follows.

$(1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2
}$

v = $sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}$

= 592999 m/s

Since, the electron is travelling downwards it means that it looses the potential energy.

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3 years ago

Is there a such thing as an elastic collision? Why do we study the case of the elastic collision in Physics?

fredd [130]

Yes there is an elastic collision in physics its when a collision occurs but no kinetic energy is loss. We study them in order to understand how to conserve momentum.

6 0

3 years ago

What quantity is scalar

daser333 [38]

A scalar quantity is a measurement of a quantity, like temperature, or mass.

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3 years ago

Read 2 more answers

A damped mass/spring system takes 14.0 s for its amplitude of the oscillator to decrease by a factor of 9. By what factor does t

fiasKO [112]

Answer:

The correct answer is "0.246".