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vova2212 [387]
3 years ago
13

Light travels at 3 × 108 m/s, and it takes about 8 min for light from the sun to travel to Earth. Based on this, the order of ma

gnitude of the distance from the sun to Earth is g
Physics
1 answer:
N76 [4]3 years ago
5 0

Answer:

The order of magnitude of the distance from the sun to Earth is 10⁸ km.

Explanation:

The order of magnitude of the distance from the sun to Earth can be calculated as follows:

c = \frac{x}{t}

Where:

c: is the speed of light = 3x10⁸ m/s

t: is the time = 8 min

Hence, the distance is:

x = c*t = 3 \cdot 10^{8} m/s*8 min*\frac{60 s}{1 min} = 1.44 \cdot 10^{11} m = 1.44 \cdot 10^{8} km

Therefore, the order of magnitude of the distance from the sun to Earth is 10⁸ km.

I hope it helps you!

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Answer:

A. 8.29\times 10^{-18}\ J

Explanation:

Given that:

p = magnitude of charge on a proton = 1.6\times 10^{-19}\ C

k = Boltzmann constant = 9\times 10^{9}\ Nm^2/C^2

r = distance between the two carbon nuclei = 1.00 nm = 1.00\times 10^{-9}\ m

Since a carbon nucleus contains 6 protons.

So, charge on a carbon nucleus is q = 6p=6\times 1.6\times 10^{-19}\ C=9.6\times 10^{-19}\ C

We know that the electric potential energy between two charges q and Q separated by a distance r is given by:

U = \dfrac{kQq}{r}

So, the potential energy between the two nuclei of carbon is as below:

U= \dfrac{kqq}{r}\\\Rightarrow U = \dfrac{kq^2}{r}\\\Rightarrow U = \dfrac{9\times10^9\times (9.6\times 10^{-19})^2}{1.0\times 10^{-9}}\\\Rightarrow U =8.29\times 10^{-18}\ J

Hence, the energy stored between two nuclei of carbon is 8.29\times 10^{-18}\ J.

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