# How would you evaluate (5.7×106 kg) × (6.3×10−2 m/s2) and express the answer in scientific notation?

Mar 29, 2016

$3.591 \times {10}^{5} N$

#### Explanation:

Probably there is a formatting issue with the question. I will attempt to be a mind-reader and restate as following

Evaluate $\left(5.7 \times {10}^{6} k g\right) \times \left(6.3 \times {10}^{-} 2 m {s}^{-} 2\right)$ and express the answer in scientific notation

$\left(5.7 \times {10}^{6} k g\right) \times \left(6.3 \times {10}^{-} 2 m {s}^{-} 2\right)$, Multiplying numbers and treating powers of $10$ separately we obtain

$\left(5.7 \times 6.3 \times {10}^{6} \times {10}^{-} 2\right) \left(k g m {s}^{-} 2\right)$
(35.91xx10^6xx10^-2))(kgms^-2),
we know that $\left(k g m {s}^{-} 2\right)$ looks like mass times acceleration and has symbol "N for Force.
Also shifting the decimal by one place to the left to represent in scientific notation
$\left(3.591 \times {10}^{1} \times {10}^{6} \times {10}^{-} 2\right) \left(N\right)$, adding powers of 10
$\left(3.591 \times {10}^{1 + 6 - 2}\right) N$
$3.591 \times {10}^{5} N$