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lesya [120]
3 years ago
15

A golf club hits a stationary 0.05kg golf ball with and average force of 5.0 x 10^3 newtons accelerating the ball at 44 meters p

er second. What is the magnitude of impulse imparted to the ball by the golf club?
Physics
1 answer:
maxonik [38]3 years ago
7 0

Answer: The magnitude of impulse imparted to the ball by the golf club is 2.2 N seconds

Explanation:

Force applied on the golf ball = 5.0\times 10^3 N

Mass of the ball = 0.05 kg

Velocity with which ball is accelerating = 44 m/s

Time period over which forece applied = t

f=ma=\frac{m\times v}{t}

t=\frac{0.05 kg\times 44m/s}{5.0\times 10^3 N}=4.4\times 10^{-4} seconds

Impulse=(force)\times (time)=f\times t = 5.0\times 10^3\times 4.4\times 10^{-4} seconds=2.2 Newton seconds

The magnitude of impulse imparted to the ball by the golf club is 2.2 N seconds

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Answer:

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Explanation:

You are missing the first part of the problem. This is an example, but it will give you the idea of how to solve yours with your data.

The first part is like this:

<em>A      4.0 cm  diameter parallel plate capacitor has a  0.44 m  m    gap. What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000 V/s?</em>

Now with this, we can solve the problem.

In order to do this, we need to use the following expression:

q = CV (1)

Where:

C: Capacitance of a parellel capacitor (in Faraday)

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However, we need is the current, and we have data of potential difference, so, all we have to do is divide the expression between time so:

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And finally, Capacitance C with two plates of area A separated by a distance d is:

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Eo = constant equals to 8.85x10^-12 F/m.

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Now, it's time to use equation (2) and solve for I:

I = 2.42x10^-11 * 500,000

I = 1.21x10^-5 A

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