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Nastasia [14]
2 years ago
9

The average velocity of a tennis ball is measured during serving and found to be 216 km/hr.

Physics
1 answer:
icang [17]2 years ago
7 0

Answer:

The time between hitting the ball and meeting the ground is t=0.33\: s

Explanation:

We have a semi parabolic motion here. We know that the initial speed 216 km/h or 60 m/s in the x-direction.

So we can use the following equation:

v=\frac{\Delta x}{\Delta t}

60=\frac{20}{t}

t=\frac{20}{60}

t=0.33\: s

hope it helps you!

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When she reaches the top, she has more potential energy than she had at the bottom.

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                                     = (40 kg) x (9.8 m/s²) x (5 m)

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3 years ago
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You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th
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<u>Answer:</u> The Young's modulus for the wire is 6.378\times 10^{10}N/m^2

<u>Explanation:</u>

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The equation representing Young's Modulus is:

Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}

where,

Y = Young's Modulus

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g = acceleration due to gravity = 9.81m/s^2

l = length of wire  = 2.6 m

A = area of cross section  = \pi r^2

r = radius of the wire = \frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m      (Conversion factor:  1 m = 1000 mm)

\Delta l = change in length  = 1.99 mm = 1.99\times 10^{-3}m

Putting values in above equation, we get:

Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2

Hence, the Young's modulus for the wire is 6.378\times 10^{10}N/m^2

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3 years ago
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Answer:

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