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Eva8 [605]
2 years ago
10

(I) A 0.145-kg baseball pitched at 31.0 m/s is hit on a horizontal line drive straight back at the pitcher at 46.0 m/s. If the c

ontact time between bat and ball is 5.00 × 10^–3 s, calculate the force (assumed to be constant) between the ball and bat.
Physics
1 answer:
tangare [24]2 years ago
8 0

Answer:

<em>The force between the ball and the bat = 2233 N</em>

Explanation:

Force: This can be defined as the product of force and its acceleration.

The S.I unit of Force is Newton (N)

F = ma .............................. Equation 1

Where F = Force , m = mass of the ball, a = acceleration of the ball.

where

a = (v-u)/t..................... Equation 2

Where v = final velocity of the ball, u = initial velocity of the ball, t = time of contact between the ball an the bat.

<em>Given: v = 46 m/s u = - 31 m/s (it hit an horizontal line drive back at the pitcher), t = 5×10⁻³ s</em>

<em>Substituting these values into equation 2,</em>

<em>a = [46-(31)]/(5×10⁻³ )</em>

a = 77<em>×10³/5</em>

<em>a = 15.4×10³ m/s².</em>

<em>Also given m = 0.145 kg</em>

<em>Substituting into equation 1,</em>

F = 0.145(15.4<em>×10³ )</em>

F = 2233 N

<em>Thus the force between the ball and the bat = 2233 N</em>

<em></em>

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Answer: Option (C) is the correct answer.

Explanation:

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2 years ago
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When a 12 V battery is connected to a 6 uF capacitor, how much energy is stored
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This has two components: a vertical and a horizontal.

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Considering the vertical motion,

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s = ut+\frac{1}{2}at^2

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1.2\text{ m} = u\sqrt{\dfrac{5.6}{g}}-\dfrac{1}{2}\times g\times\dfrac{5.6}{g}

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Answer:

Explanation:

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E_1=E_2

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