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Ivahew [28]
3 years ago
14

neptune is an average distance of 4.5×10^12m from the sun. Estimate the length of the Neptunian year.

Physics
1 answer:
Vikentia [17]3 years ago
6 0

As per Kepler's third law we know that

\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}

now here we know that

T_1 = year of Neptune

T_2 = year of Earth

R_1 = distance of Neptune from Sun

R_2 = Distance of Earth from Sun

so now we will have

\frac{T_1^2}{1} = \frac{(4.5 \times 10^{12})^3}{(1.5 \times 10^11)^3}

T_1^2 = 27000

T_1 = 164.3 years

so length of year of Neptune is 164.3 years

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5.7 x 10^12 C

Explanation:

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the gravitational force between them is

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The electrostatic force between them is

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According to the question

1 % of Fg = Fe

0.01 \times 6.67\times10^{-11}\frac{5.97 \times 10^{24}\times7.35 \times 10^{22}}{d^{2}}=9 \times 10^{9}\frac{q^{2}}{d^{2}}

2.927 \times 10^{35}=9 \times10^{9}q^{2}

3.25 \times 10^{25}=q^{2}

q = 5.7 x 10^12 C

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2 years ago
Infer why the output force exerted by a rake must be less than input force?
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Answer:

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where the potential is

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Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching

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