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how would you calculate the kinetic energy of a 0.148 kg (5.22 oz) base…

Question

how would you calculate the kinetic energy of a 0.148 kg (5.22 oz) baseball traveling at 40 m/s (90 mph)? (0.148)(40) (0.148)(40)9.8 (0.148)(40)2 9.8 (0.148)(40)2

Explanation:

Step1: Recall kinetic - energy formula

The formula for kinetic energy is $K = \frac{1}{2}mv^{2}$, where $m$ is the mass and $v$ is the velocity.

Step2: Identify values of $m$ and $v$

Given $m = 0.148\ kg$ and $v = 40\ m/s$.

Step3: Substitute values into formula

$K=\frac{1}{2}(0.148)(40)^{2}$.

Step4: Calculate the result

First, $(40)^{2}=1600$. Then, $\frac{1}{2}(0.148)(1600)=(0.148)(800) = 118.4\ J$.

Answer:

The kinetic - energy of the baseball is calculated using the formula $K=\frac{1}{2}mv^{2}$, substituting $m = 0.148\ kg$ and $v = 40\ m/s$ gives $K=(0.148)(40)^{2}\div2$. So the correct expression for calculating the kinetic energy is $(0.148)(40)^{2}\div2$.