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3 years ago

10

Heather drops a ball weighing 0.50kg a distance of 2.0m. If the speed of the ball right after it bounces is the same speed of th

e ball right before it bounces, and the interaction time between the ball and the floor was 0.10 seconds, what was the average force of the floor on the ball?

1 answer:

ki77a [65]3 years ago

7 0

The speed of the ball just before impact was v=√(2gh) = 6.26m/s. The acceleration is twice this over the time (twice because the second speed is the same in the other direction, meaning the total change in speed is 2V)
a = 12.52/0.10 = 125.2m/s²
The force is F=ma, so F = 0.5kg·125.2m/s² = 62.6N

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The drag coefficient is $D_z = 1.30512$

Explanation:

From the question we are told that

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let assume that the velocity is 20 m/s

Then

$F_{\alpha } = 0.9226 * 20^{0.5737}$

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Where

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$\rho$ is the density of the fluid

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substituting values

$A = 3.142 * \frac{(0.15)^2}{4}$

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Then

$D_z = \frac{2 * 5.1453 }{0.0176 * 1.12 * 20^2 }$

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2 years ago

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Answer:

a) I₁ = 11.2 Lux , vertical direction , b) I₂ = 1.44 Lux

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is polarized in the vertical direction

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Answer:

The time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$

Explanation:

Given :

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(A)

First find the capacitance of capacitor,

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Where $\epsilon _{o} = 8.85 \times 10^{-12}$

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Where $Q = It$

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Now differentiate above equation wrt. time,

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3 years ago

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3 years ago

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Answer: The original temperature was

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$PV=nRT$

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Now:

$T_{1}=\frac{P_{1}V_{1}}{nR}$

⇒ $T_{1}=126.51K$

7 0

2 years ago

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