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Amanda [17]
3 years ago
6

A(n) 0.47 kg softball is pitched at a speed of 10 m/s. The batter hits it back directly at the pitcher at a speed of 29 m/s. The

bat acts on the ball for 0.019 s. What is the magnitude of the impulse imparted by the bat to the ball? Answer in units of N · s.
Physics
1 answer:
nadya68 [22]3 years ago
8 0

Answer:

18.33 Ns

Explanation:

As the pitch back speed has the opposite direction as before, the change in velocity would be

\Delta v = v_2 - v_1 = 29 - (-10) = 39 m/s

So the change in momentum of the ball would be the product of its velocity change and its mass

\Delta p = \Delta v m = 39 * 0.47 = 18.33 kgm/s

This is equals to the impulse acted on the ball by the bat, which is 18.33 Ns

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Answer:

(a) Approximately 10.1\; {\rm V}.

Explanation:

Let C denote the capacitance of a capacitor. Let V be the potential difference (voltage) between the two plates of this capacitor. The energy E stored in this capacitor would be:

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Rearrange this equation to find an expression for the potential difference V in terms of capacitance C and energy E:

\begin{aligned}V^{2} &= \frac{2\, E}{C} \end{aligned}.

\begin{aligned}V &= \sqrt{\frac{2\, E}{C}} \end{aligned}

The capacitance C of this capacitor is given in nanofarads. Convert that unit to standard unit (farads):

\begin{aligned}C &= 360\; {\rm nF} \\ &= 360\; {\rm nF} \times \frac{1\; {\rm F}}{10^{9}\; {\rm nF}} \\ &= 3.60 \times 10^{-7}\; {\rm F}\end{aligned}.

Given that the energy stored in this capacitor is E = 1.85 \times 10^{-5}\; {\rm J}, the potential difference across the capacitor plates would be:

\begin{aligned}V &= \sqrt{\frac{2 \times 1.85 \times 10^{-5}\; {\rm J}}{3.60 \times 10^{-7}\; {\rm F}}} \\ &\approx 10.1\; {\rm V}\end{aligned}.

7 0
2 years ago
Ehren is trying to increase his mile run from eight to six minutes. He has decided to run some sandy hills near his home. Which
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timama [110]

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