the magnitude of the electrical force acting between a +2.4×10⁻⁸ c charge and a +1.8×10⁻⁶ c charge that are…

Explanation:

Step1: Identify the formula

Use Coulomb's law $F = k\frac{q_1q_2}{r^2}$, where $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, $q_1 = 2.4\times 10^{-8}\ C$, $q_2=1.8\times 10^{-6}\ C$, and $r = 0.008\ m$.

Step2: Substitute the values

$F=9\times 10^{9}\times\frac{(2.4\times 10^{-8})\times(1.8\times 10^{-6})}{(0.008)^{2}}$ First, calculate the numerator: $(2.4\times 10^{-8})\times(1.8\times 10^{-6})=2.4\times1.8\times10^{-8 - 6}=4.32\times 10^{-14}$. Then, calculate the denominator: $(0.008)^{2}=6.4\times 10^{-5}$. So, $F = 9\times 10^{9}\times\frac{4.32\times 10^{-14}}{6.4\times 10^{-5}}$.

Step3: Simplify the expression

$F=\frac{9\times4.32\times 10^{9-14}}{6.4\times 10^{-5}}=\frac{38.88\times 10^{-5}}{6.4\times 10^{-5}}$. Since $\frac{10^{-5}}{10^{-5}} = 1$, then $F=\frac{38.88}{6.4}=6.075$.

Step4: Round the result

Rounding $6.075$ to the tenths place gives $6.1$.

Answer:

$6.1$