Ask question

Ask question

All categories

3 years ago

8

Imagine a Rip-van-Winkle type who lives in the mountains. Just before going to sleep he yells "WAKE UP" and the sound echoes off

the nearest mountain and returns 8 hours later.
How far away is that mountain? The speed of sound is __ m/s.

1 answer:

stiv31 [10]3 years ago

6 0

Answer:

4939200 m

Explanation:

v = Velocity of sound in air = 343 m/s (general value)

For the echo to reach Rip van Winkle it must cover the distance to the mountain twice. So, time taken for the sound traveling to the mountain once

$t=\dfrac{8}{2}=4\ hr$

Distance is given by

$s=vt\\\Rightarrow s=343\times 4\times 60\times 60\\\Rightarrow s=4939200\ m=4939.2\ km$

The distance to the mountain is 4939200 m

You might be interested in

PLS HELP DUE IN 30 MINS!!! True or false: A large-amplitude pulse travels at a faster speed than a small-amplitude pulse. Explai

OLga [1]

Answer:

False

Explanation:

Larger pulse with more energy means that the wave has large amplitude. As in the given question, both the waves are travelling in the same medium, hence their speeds will remain same and therefore the larger pulse will not overtake the smaller pulse. Remember the amplitude of a wave does not affect the speed at which the wave travels

8 0

2 years ago

Read 2 more answers

Glass absorbs ultraviolet (UV) rays from the Sun. Would a fraction of the incident UV light be reflected from the air/glass boun

fiasKO [112]

Answer:

Yes

Explanation:

Any transparent surface in practical is neither a perfect absorber of electromagnetic waves neither a perfect reflector. Generally all the transparent surfaces reflect some amount of irradiation and the other parts are absorbed and transmitted.

<u>That is given by as relation:</u>

$\alpha+\rho+\tau=1$

where:

$\alpha=$ absorptivity which is defined as the ratio of the absorbed radiation to the total irradiation

$\rho=$ reflectivity is defined as the ratio of reflected radiation to the total irradiation

$\tau=$ transmittivity is defined as the ratio of total transmitted radiation to the total irradiation

6 0

3 years ago

Calculate the energy needed to heat 4 kg of water from 25°C to 45°C.

inna [77]

(1 cal/g °C) x (4000 g) x (45 - 25)°C = 80000 cal = 80 kcal. So the answer is 80 kcal .

8 0

3 years ago

Read 2 more answers

A runner is moving at a constant speed of 8.00 m/s around a circular track. If the distance from the runner to the center of the

Genrish500 [490]

Answer: Last option

2.27 m/s2

Explanation:

As the runner is running at a constant speed then the only acceleration present in the movement is the centripetal acceleration.

If we call a_c to the centripetal acceleration then, by definition

$a_c =w^2r = \frac{v^2}{r}$

in this case we know the speed of the runner

$v =8.00\ m/s$

The radius "r" will be the distance from the runner to the center of the track

$r = 28.2\ m$

$a_c = \frac{8^2}{28.2}\ m/s^2$

$a_c = 2.27\ m/s^2$

The answer is the last option

3 0

3 years ago

Carbon is allowed to diffuse through a steel plate 9.7-mm thick. The concentrations of carbon at the two faces are 0.664 and 0.3

cupoosta [38]

Answer:

844°C

Explanation:

The problem can be easily solve by using Fick's law and the Diffusivity or diffusion coefficient.

We know that Fick's law is given by,

$J = - D \frac{\Delta c}{\Delta x}$

Where $\frac{\Delta c}{\Delta x}$ is the concentration of gradient

D is the diffusivity coefficient

and J is the flux of atoms.

In the other hand we have, that

$D= D_0 e^{\frac{E_d}{RT}}$

Where $D_0$ is the proportionality constant,

R is the gas constant, T the temperature and $E_d$ is the activation energy.

Replacing the value of diffusivity coefficient in Fick's law we have,

$J = -D_0 ^{\frac{E_d}{RT}}\frac{\Delta c}{\Delta x}$

Rearrange the equation to get the value of temperature,

$T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}$

We have all the values in our equation.

$\Delta c = 0.664-0.339 = 0.325 C. cm^{-1}$

$\Delta x = 9.7*10^{-3}m$

$E_d = 82000J$

$D_0 = 6.5*10^{-7}m^2/s$

$J = 3.2*10^{-9}m^2/s$

$R= 8.31Jmol^{-1}K$

Substituting,

$T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}$

$T=-\frac{-82000}{(8.31)ln(\frac{3.2*10^{-9}(9.7*10^{-3})}{6.5*10^{-7} (0.325)})}$

$T=1118.07K=844\°C$

4 0

3 years ago