Answer:
30 kg·m/s
Explanation:
To address this question, we'll apply the principle of impulse and momentum. The impulse experienced by an object is equal to the change in its momentum, and this can be calculated using the formula:
[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Impulse-Momentum Theorem:}}\\\\\vec I=\Delta\vec p =\vec F_{avg}\Delta t \\ \\ \text{Where:}\\\bullet \ \vec I \text{ represents the impulse}\\ \bullet \ \Delta \vec p \text{ represents the change in momentum}\\\bullet \ \vec F_{avg} \text{ represents the force applied}\\\bullet \ \Delta t\text{ represents the change in time}\end{array}\right}[/tex]
We are given:
Plug in our given values and solve for 'Δp':
[tex]\Longrightarrow \Delta\vec p =(300 \text{ N})(0.1\text{ s})\\\\\\\\\therefore \Delta\vec p = \boxed{30 \ \frac{\text{kg $\cdot$ m}}{s}}[/tex]
Thus, the change in momentum of the baseball is 30 kg·m/s.