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Vinvika [58]
2 years ago
12

How is a light element like helium is formed in stars and how is a heavier element like gold formed by stars?

Physics
1 answer:
aliina [53]2 years ago
7 0

Answer:

C

Explanation:

Because *censored*. Then the element *censored* your answer

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When nuclear fission occurs some mass is lost where does that mass go
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The loss of matter is called the mass defect. The missing matter is converted into energy. You can actually calculate the amount of energy produced during a nuclear reaction with  fairly simple equation developed by Albert Einstein; E = mc^2. In this equation, E is the amount of energy produced, m is the missing mass, or the mass defect, and c is the speed of light, which is a rather large number. The speed of light is squared, making that part of the equation a very large number that, even when multiplied by a small amount of mass, yields a large amount of energy.
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True or False: A cross product is an equation which states that two ratios are equal.
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At what distance from Earth does the force of gravity exerted by Earth on the coasting spacecraft cancel the force of gravity ex
telo118 [61]

Answer:

346 * 10⁶ m

Explanation:

The force of gravity of the earth that will cancel the the force of gravity exerted by the moon will be equal to each other

Let F_{e} be the force of gravity exerted by the earth

and let F_{m} be the force of gravity exerted by the moon

According to Newton's law of universal gravitation, the force of attraction between two different masses, m₁ and m₂ separated by a distance, d,  is given by:

F = \frac{Gm_{1} m_{2} }{d^{2} }

Mass of the earth, m_{e} = 5.97 * 10^{24} kg

Mass of the moon, m_{m} = 7.348 * 10^{22} kg

Mass of the satellite, m_{s} = ?

F_{e}  = \frac{G*5.97 * 10^{24} M }{d^{2} }...............................(1)

The earth and the moon are separated by a distance, 3.844 * 10⁸ m

F_{m}  = \frac{G*7.348 * 10^{22} M }{(3.844 * 10^{8} - d) ^{2} }............................(2)

Equating equations (1) and (2)

\frac{5.97 * 10^{24} }{d^{2} } = \frac{7.348 * 10^{24} }{(3.844* 10^{8} -d)^{2} }

(5.97 * 10^{24})(14.78 * 10^{16}  -7.688*10^{8}d + d^{2}) = 7.348 * 10^{24} d^{2} \\88.24*10^{40} - 45.9 * 10^{32}d +  5.97 * 10^{24}d^{2} =  7.348 * 10^{24} d^{2}\\ 1.378 * 10^{24}d^{2} + 45.9 * 10^{32}d + 88.24*10^{40} = 0\\

Factorising out 10^{24}

1.378d^{2} + 45.9 * 10^{8}d + 88.24*10^{16} = 0

Solving for d in the quadratic equation  above:

d = 346 * 10⁶ m

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It is the atmosphere

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If the velocity of the particle is uniform, then the path of the particle must be a​
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