To solve this problem, we will go through parts (a) and (b) step by step.
(a) Efficiency of the Energy Transformation
The efficiency \(\eta\) of an energy transformation can be calculated using the formula:
\[ \eta = \frac{\text{useful energy output}}{\text{total energy input}} \times 100% \]
Here, the useful energy output is the energy transferred to the baseball (70 J), and the total energy input is the initial kinetic energy of the bat (120 J).
Now, substituting the values:
\[ \eta = \frac{70 \text{ J}}{120 \text{ J}} \times 100% \] \[ \eta = \frac{70}{120} \times 100% \] \[ \eta = 0.5833 \times 100% \] \[ \eta \approx 58.33% \]
Thus, the efficiency of the energy transformation is approximately 58.33%.
(b) Speed of the Baseball Immediately After It Has Been Hit
To find the speed of the baseball immediately after being hit, we can use the formula for kinetic energy:
\[ KE = \frac{1}{2}mv^2 \]
where
- \(KE\) is the kinetic energy,
- \(m\) is the mass of the baseball, and
- \(v\) is the speed of the baseball.
We know the kinetic energy transferred to the baseball is 70 J and the mass \(m\) of the baseball is 0.145 kg. We can rearrange the formula to solve for \(v\):
\[ 70 = \frac{1}{2} \times 0.145 \times v^2 \]
First, multiply both sides by 2:
\[ 140 = 0.145 \times v^2 \]
Now, divide both sides by 0.145:
\[ v^2 = \frac{140}{0.145} \] \[ v^2 \approx 965.5172 \]
Now, take the square root of both sides to find \(v\):
\[ v \approx \sqrt{965.5172} \approx 31.1 \text{ m/s} \]
Thus, the speed of the baseball immediately after it has been hit is approximately 31.1 m/s.
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