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Semmy [17]
2 years ago
10

You yell into a canyon. You hear the echo in 3 seconds. How long did the sound of your voice travel before bouncing off a cliff?

Physics
1 answer:
Lilit [14]2 years ago
7 0

Answer:

The sound travelled 516 meters before bouncing off a cliff.

Explanation:

The sound is an example of mechanical wave, which means that it needs a medium to propagate itself at constant speed. The time needed to hear the echo is equal to twice the height of the canyon divided by the velocity of sound. In addition, the speed of sound through the air at a temperature of 20 ºC is approximately 344 meters per second. Then, the height of the canyon can be derived from the following kinematic formula:

2\cdot h = v\cdot t (1)

Where:

h - Height, measured in meters.

v - Velocity of sound, measured in meters per second.

t - Time, measured in seconds.

If we know that  v = 344\,\frac{m}{s} and t = 3\,s, then the height of the canyon is:

h = \frac{v\cdot t}{2}

h = \frac{\left(344\,\frac{m}{s} \right)\cdot (3\,s)}{2}

h = 516\,m

The sound travelled 516 meters before bouncing off a cliff.

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Our milky way galaxy is 100000 lyly in diameter. a spaceship crossing the galaxy measures the galaxy's diameter to be a mere 1.
Sidana [21]

The speed of the spaceship relative to the galaxy is 0.99999995c.

A light-year measures distance rather than time (as the name might imply). A light-year is a distance a light beam travels in one year on Earth, which is roughly 6 trillion miles (9.7 trillion kilometers). One light-year equals 5,878,625,370,000 miles. Light moves at a speed of 670,616,629 mph (1,079,252,849 km/h) in a vacuum.We multiply this speed by the number of hours in a year to calculate the distance of a light-year (8,766).

The Milky way galaxy is 100,000 light years in diameter.

The galaxy's diameter is a mere 1. 0 ly.

We know that ;

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L₀ = 100,000 light year

1 = 100,000 \sqrt{1-\frac{v^2}{c^2} }

1 = 100,000 \sqrt{1-\frac{v^2}{(3*10^8)^2} }

\frac{1}{100,000}  = \sqrt{1-\frac{v^2}{c^2} }

v = 0.999999995 c

Therefore, the speed of the spaceship relative to the galaxy is 0.99999995c.

Learn more about a light year here:

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5 0
1 year ago
Imagine an alternate universe where the value of the Planck constant is . In that universe, which of the following objects would
HACTEHA [7]

Question: The planck constant was not given. In this calculation, planck constant of 6.62607*10^-9 Js  is used for the calculation.

Answer:

(a) A virus -------------Classical

(b) A buckyball -----Classical

(c) A mosquito ------ Quantum

(d) A turtle  ------------Quantum

Explanation:

 Calculating the wavelength using the formula;

λ= h/(mv)

where

λ= Wavelength

h = Planck Constant = 6.62607*10^-9 Js

m = mass in kg

v = velocity in m/s

Virus size = 280. nm = 2.80*10⁻⁷ m

a)

A Virus:

m = 9.4 x 10-17 g 9.4*10⁻²⁰ kg

v = 0.50 µm/s = 5 *10⁻⁷ m/s

h = 6.62607*10^-9 Js

Virus size = 280 nm = 2.80*10⁻⁷ m

Substituting into the formula; we have

λ= h/(mv)

λ= 6.62607*10^-9/ (9.4*10⁻²⁰* 5 *10⁻⁷)

  = 6.62607*10^-9/4.7*10^-26

  = 1.4*10^17 m

Classical : Wavelength is bigger than it's size

(b)

A buckyball

m = 1.2 x 10-21 g = 1.2 *10⁻²⁴ kg

V = 37 m/s

Size = 0.7 nm = 7*10⁻¹⁰ m

Substituting into the formula, we have

λ= h/(mv)

λ= 6.62607*10^-9/ ( 1.2 *10⁻²⁴* 37)

  =  6.62607*10^-9/4.44*10^-23

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Classical : Wavelength is bigger than it's size

(c)

A mosquito

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v = 1.1 m/s

Size =  6.3 mm = 6.3*10⁻³ m

Substituting into the formula, we have

λ= h/(mv)

λ= 6.62607*10^-9/ (  1*10⁻⁶* 1.1)

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Quantum Approach: The wavelength and the size are comparable

(d)

A turtle

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Size =  22. cm = 0.22 m

V =  2.8 cm/s. = 0.028 m/s

Substituting into the formula, we have

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8 0
3 years ago
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inessss [21]

Answer:

Option C

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