elhadaverdeibd7

2023-02-22

If the distance between two charges is doubled, what will happen the force between the charges?

ri2n1tv9pt

Beginner2023-02-23Added 8 answers

The right decision is B 0.248

We have operation is $\frac{1}{5}+\frac{1}{25}+\frac{1}{125}$

Using the denominators as factors,

$\frac{1}{5}+\frac{1}{25}+\frac{1}{125}=\frac{1}{5}+\frac{1}{{5}^{2}}+\frac{1}{{5}^{3}}$

To change a fraction into decimal form, make the denominator in terms of multiples of 10.

$\Rightarrow (\frac{1}{5}\times \frac{2}{2})+(\frac{1}{{5}^{2}}\times \frac{{2}^{2}}{{2}^{2}})+(\frac{1}{{5}^{3}}\times \frac{{2}^{3}}{{2}^{3}})$

$=\frac{2}{10}+\frac{4}{100}+\frac{8}{1000}$

= 0.248

We have operation is $\frac{1}{5}+\frac{1}{25}+\frac{1}{125}$

Using the denominators as factors,

$\frac{1}{5}+\frac{1}{25}+\frac{1}{125}=\frac{1}{5}+\frac{1}{{5}^{2}}+\frac{1}{{5}^{3}}$

To change a fraction into decimal form, make the denominator in terms of multiples of 10.

$\Rightarrow (\frac{1}{5}\times \frac{2}{2})+(\frac{1}{{5}^{2}}\times \frac{{2}^{2}}{{2}^{2}})+(\frac{1}{{5}^{3}}\times \frac{{2}^{3}}{{2}^{3}})$

$=\frac{2}{10}+\frac{4}{100}+\frac{8}{1000}$

= 0.248

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