Question

$\frac{(- 8.99\times10^{9})(15\times10^{-6}c)(11\times10^{-6}c)}{(0.125)^{2}}\timescos(37^{circ})$

Explanation:

Step1: Calculate the product in the numerator

First, calculate \((- 8.99\times10^{9})(15\times10^{-6})(11\times10^{-6})\).
\[(-8.99\times10^{9})\times(15\times10^{-6})\times(11\times10^{-6})=(-8.99\times15\times11)\times(10^{9}\times10^{-6}\times10^{-6})\]
\[=(-8.99\times15\times11)\times10^{9 - 6-6}=(-8.99\times165)\times10^{-3}\]
\[=-1483.35\times10^{-3}=-1.48335\]

Step2: Calculate the denominator

Calculate \((0.125)^{2}=0.125\times0.125 = 0.015625\)

Step3: Divide the numerator by the denominator

\(\frac{-1.48335}{0.015625}=-94.936\)

Step4: Multiply by \(\cos(37^{\circ})\)

Since \(\cos(37^{\circ})\approx0.7986\), then \(-94.936\times0.7986\approx - 75.72\)

Answer:

\(-75.72\)