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3 years ago

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A balloon is expanded to the same volume as that of a human head. Do an order-of-magnitude estimate of the volume of this balloo

n in cm^3
a. 100cm^3

b. 1,000,000cm^3

c. 10cm^3

d. 1,000cm^3

1 answer:

Grace [21]3 years ago

4 0

Answer:

Volume of head = 1000 cm^3

Explanation:

The head of a normal person can be assumed as a sphere with radius 10 cm.

Volume of sphere $=\frac{4}{3} \pi r^3$ , where r is the radius.

We have approximate radius = 10 cm.

Approximate volume of head $=\frac{4}{3}* \pi *10^3=4189 cm^3$

In the given options the closest value to the approximate volume is 1000 cm^3.

So, volume of head = 1000 cm^3

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Answer:

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Solution:

As per the question:

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Thickness of the tennis ball, t = 2.0 mm = $2.0\times 10^{- 3}\ m$

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Now,

The maximum kinetic energy of the tennis ball is given by:

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Now, the distance the ball can penetrate to is given by:

$\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}$

$\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js$

Thus

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

$\eta = 1.52\times 10^{-35}\ m$

Now,

We can calculate the tunneling probability as:

$P(t) = e^{\frac{- 2t}{\eta}}$

$P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}$

$P(t) = e^{-2.63\times 10^{32}}$

Taking log on both the sides:

$logP(t) = -2.63\times 10^{32} loge$

$P(t) = 10^{- 1.17\times 10^{32}}$

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Answer:

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3 years ago

Read 2 more answers

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pshichka [43]

Answer:

The average speed for the entire run is 12 km/h.

Explanation:

The average speed is given by the following equation:

$\overline{v} = \frac{d_{T}}{t_{T}}$

Where:

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$t_{T}$ : is the total time

If during the first hour, they ran a total of 13 kilometers and then, they ran 5.0 kilometers during the next half an hour we have:

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I hope it helps you!

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3 years ago