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

Explanation:

Step1: Identify Coulomb's law formula

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

Step2: Substitute values into formula

$F=(9\times 10^{9})\frac{(2.4\times 10^{-6})(1.8\times 10^{-6})}{(0.008)^{2}}$ First, calculate the numerator: $(2.4\times 10^{-6})(1.8\times 10^{-6})=2.4\times1.8\times10^{-6 - 6}=4.32\times 10^{-12}$. Then, calculate the denominator: $(0.008)^{2}=6.4\times 10^{-5}$. So, $F=(9\times 10^{9})\frac{4.32\times 10^{-12}}{6.4\times 10^{-5}}$.

Step3: Simplify the expression

$\frac{4.32\times 10^{-12}}{6.4\times 10^{-5}}=\frac{4.32}{6.4}\times10^{-12 + 5}=0.675\times 10^{-7}=6.75\times 10^{-8}$. $F=(9\times 10^{9})\times(6.75\times 10^{-8})$. Using the rule of exponents $a^m\times a^n=a^{m + n}$, we have $F = 9\times6.75\times10^{9-8}=60.75\ N$.

Step4: Round to the tenths place

$F\approx60.8\ N$

Answer:

$60.8$