Question

I need help asap! Will give brainliest

Answer 1

$\\ \tt\hookrightarrow \dfrac{(3\times 10^4kg)(4\times 10^4m)}{6\times 10^4s)}$

$\\ \tt\hookrightarrow \dfrac{12\times 10^8kgm}{6\times 10^4s}$

$\\ \tt\hookrightarrow 2kgm/s$

We have calculated momentum