3 years ago

75000×(151÷100)².WHAT IS ITS VALUE ???

Physics

1 answer:

Lady_Fox [76]3 years ago

3 0

Answer:

171007 1/2

Explanation:

it's a mixed number

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What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of −25.0 nc? g?

Daniel [21]

To calculate the force between two negative charges, we use the formula which is given by the Coulomb`s Law as

$F=\frac{kq_{1}q_{2} }{r^{2} }$

Here, $q_{1}$ and $q_{2}$ are the charges on the pith balls, r is the separation between the charges and k is constant and its value is $8.99\times 10^9 N m^2/C^2$ .

Given $q_{1} =q_{2} =-25 nC=-25\times 10^{-9}C$ and $r=8 cm=8\times10^{-2} m$ .

Substituting these values in above formula we get,

$F= 8.99\times 10^9 N m^2/C^2\frac{(-25\times 10^{-9}C)(-25\times 10^{-9}C)}{(8\times10^{-2} m)^2} \\\\\ F= 877929.7\times 10^{-9} N\\\\F=8.8\times10^{-4}N$