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2 years ago

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A common acoustic phenomenon is thunder, the rapid heating and supersonic expansion of air created by a lightning stroke. Becaus

e the speed of light reaches almost any point on Earth instantly while sound travels more slowly, the time lag between lightning and thunder can be used to estimate distance to the lightning. This delay time is ______ seconds per kilometer (5 sec/mi).

1 answer:

Oksana_A [137]2 years ago

4 0

Answer:

Δt ≈ 2.9137 ≈ 3 seconds per kilometer

Explanation:

The speed of light is approximately 300000 km /s, while the speed fo the sound in the air is 0.3432 km/s.

The light takes therefore this time to travel one kilometer

$x = vt$ $t = \frac{x}{v}$

$t = \frac{1}{300000}$

$t = 3.33 x 10^{-6} s$

On the other hand the sound takes this time to travel one kilometer

$t = \frac{1}{0.3432}$

t = 2.9137 s

Then the delay time is 2.9137 - $3.33 x 10^{-6} s$

Δt ≈ 2.9137 ≈ 3 s

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The current theory of the structure of the

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Answers:

a) $2.82(10)^{21} kg$

b) $1410 J$

c) $36.62 m/s$

Explanation:

<h3>a) Mass of the continent</h3>

Density $\rho$ is defined as a relation between mass $m$ and volume $V$ :

$\rho=\frac{m}{V}$ (1)

Where:

$\rho=2720 kg/m^{3}$ is the average density of the continent

$m$ is the mass of the continent

$V$ is the volume of the continent, which can be estimated is we assume it as a a slab of rock 5300 km on a side and 37 km deep:

$V=(length)(width)(depth)=(5300 km)(5300 km)(37 km)=1,030,330,000 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=1.03933(10)^{18} m^{3}$

Finding the mass:

$m=\rho V$ (2)

$m=(2720 kg/m^{3})(1.03933(10)^{18} m^{3})$ (3)

$m=2.82(10)^{21} kg$ (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy $K$ is given by the following equation:

$K=\frac{1}{2}mv^{2}$ (5)

Where:

$m=2.82(10)^{21} kg$ is the mass of the continent

$v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1(10)^{-9} m/s$ is the velocity of the continent

$K=\frac{1}{2}(2.82(10)^{21} kg)(1(10)^{-9} m/s)^{2}$ (6)

$K=1410 J$ (7) This is the kinetic energy of the continent

<h3>c) Speed of the jogger</h3>

If we have a jogger with mass $m=77 kg$ and the same kinetic energy as that of the continent $1413 J$ , we can find its velocity by isolating $v$ from (5):

$v=\sqrt{\frac{2 K}{m}}$ (6)

$v=\sqrt{\frac{2 (1413 J)}{77 kg}}$

Finally:

$v=36.62 m/s$ This is the speed of the jogger

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alekssr [168]

Answer:

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Explanation:

and 27.1 secs after

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Answer:

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