I would say your answer to this question would be D
We will have the following:
![\begin{gathered} Q=mc\Delta T\Rightarrow Q=(36)(4.18)(48-24) \\ \\ \Rightarrow Q=3611.52 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20Q%3Dmc%5CDelta%20T%5CRightarrow%20Q%3D%2836%29%284.18%29%2848-24%29%20%5C%5C%20%20%5C%5C%20%5CRightarrow%20Q%3D3611.52%20%5Cend%7Bgathered%7D)
So, the heat to add is 3611.52 Joules.
I think it would be the scientific method.
Answer:
46.45 m/s
Explanation:
Total momentum before jump = Total momentum after jump
11.7 * ( 36.7 + 45.2 ) = ( 36.7 * (-31.1) ) + ( 45.2 * v )
v*45.2 - 1141.37 = 958.23
v = 2099.6/45.2 = 46.45