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arsen [322]
3 years ago
9

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).
Physics
1 answer:
Oksana_A [137]3 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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It should be A

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graph A

Explanation:

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but the slope for the velocity time graph is acceleration, acceleration is vector quantity ( has magnitude and direction)

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Ignoring air resistance and the little friction from the plastic tube, the magnet was a freely-falling object in each trial. If
horsena [70]

Answer:

emf will also be 10 times less as compared to when it has fallen 40 \mathrm{m}

Explanation:

We know, from faraday's law-

e m f=-N \frac{\Delta \Phi}{\Delta T}

and \Phi=B . A

So, as the height increases the velocity with which it will cross the ring will also increase. (v=\sqrt{2 g h})

Given

\mathrm{V} 1(\text { Speed at } 40 \mathrm{m})=2 \mathrm{x} \mathrm{V} 2(\text { speed at } 10 \mathrm{m})

\sqrt{2 g h_{2}}=2 \times \sqrt{2 g h_{1}}=28.28 \mathrm{m} / \mathrm{s}

Now, from 40 \mathrm{cm}

V_{3}=\sqrt{2 g h_{3}}=\sqrt{2 \times 10 \times 0.4}=2.82 \mathrm{m} / \mathrm{s}

From equation a and b we see that velocity when dropped from 40 \mathrm{m} is 10 times greater when height is 40 \mathrm{cm} so, emf will also be 10 times less as compared to when it has fallen 40 \mathrm{m}

4 0
3 years ago
Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce
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Answer:

speed of electrons = 3.25 × 10^{7} m/s

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radius of circular orbit is 2.76 × 10^{-4} m

Explanation:

given data

voltage = 3 kV

magnetic field = 0.66 T

solution

law of conservation of energy

PE = KE

qV = 0.5 × m × v²

v = \sqrt{\frac{2qV}{m}}

v = \sqrt{\frac{2\times 1.6 \times 10^{-19}\times 3}{9.1\times 10^{-31}}

v = 3.25 × 10^{7} m/s

and

magnetic force on particle movie in magnetic field

F = Bqv

ma = Bqv

a = \frac{Bqv}{m}  

a =  \frac{0.67\times 1.6\times 10^{-19}\times 3.25\times 10^7}{9.1\times 10^{-31}}

a = 3.82 × 10^{18} m/s²

and acceleration in term g

a = \frac{3.82\times 10^{18}}{9.81}  

a = 3.9 × 10^{17} g

acceleration in term g is 3.9 × 10^{17} g.

and

electron moving in circular orbit has centripetal force

F = \frac{mv^2}{r}  

Bqv = \frac{mv^2}{r}  

r = \frac{mv}{Bq}  

r = \frac{9.1\times 10^{-31}\times 3.25\times 10^7}{0.67\times 1.6\times 10^{-19}}  

r = 2.76 × 10^{-4} m

radius of circular orbit is 2.76 × 10^{-4} m

8 0
4 years ago
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