Ask question

Ask question

All categories

anastassius [24]

2 years ago

14

A person hears the echo of his own voice from a distant hill after 2 seconds. How far away is the person from the hill, if the s

peed of sound is 330m/s ?

1 answer:

olga nikolaevna [1]2 years ago

4 0

Answer: 330 m

Explanation:

The speed of sound $V$ is defined as the distance traveled by the sound wave $d$ in a especific time $t$ :

$V=\frac{d}{t}$

Where:

$V=330 m/s$ is the speed of sound

$t=\frac{2 s}{2}=1 s$ is half the time the sound wave travels since the person speaks, the sound wave hits the hill and then returns to the person again as an echo

$d$ is the distance between the person and the hill

So, we have to find $d$ :

$d=Vt$

$d=(330 m/s)(1 s)$

Finally:

$d=330 m$

You might be interested in

Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ

bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; $H_0 = 51km/s/Mly$

Age of the universe; $t = \ ?$

We know that, the reciprocal of the Hubble's constant ( $H_0$ ) gives an estimate of the age of the universe ( $t$ ). It is expressed as:

$Age\ of\ Universe; t = \frac{1}{H_0}$

Now,

Hubble's constant; $H_0 = 51km/s/Mly$

We know that;

$1\ light\ years = 9.46*10^{15}m$

so

$1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m$

Therefore;

$H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly} \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 = 5.39 *10^{-18}s^{-1}\\$

Now, we input this Hubble's constant value into our equation;

$Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t = \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years$

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

6 0

2 years ago

I NEED THIS FAST!

Romashka [77]

The variable you can change in an experiment is ( B ) A dependent variable

6 0

2 years ago

Read 2 more answers

A particular coaxial cable is comprised of inner and outer conductors having radii 1 mm and 3 mm respectively, separated by air.

noname [10]

Answer:

The value is $\rho_s = 4.026 *10^{-6} \ C/m^2$

Explanation:

From the question we are told that

The radius of the inner conductor is $r_1 = 1 \ mm = 0.001 \ m$

The radius of the outer conductor is $r_2 = 3 \ mm = 0.003 \ m$

The potential at the outer conductor is $V = 1.5 kV = 1.5 *10^{3} \ V$

Generally the capacitance per length of the capacitor like set up of the two conductors is

$C= \frac{2 * \pi * \epsilon_o }{ ln [\frac{r_2}{r_1} ]}$

Here $\epsilon_o$ is the permitivity of free space with value $\epsilon_o = 8.85*10^{-12} C/(V \cdot m)$

=> $C= \frac{2 * 3.142 * 8.85*10^{-12} }{ ln [\frac{0.003}{0.001} ]}$

=> $C= 50.6 *10^{-12} \ F/m$

Generally given that the potential of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge

Generally the line charge density of the outer conductor is mathematically represented as

$\rho_l = C * V$

=> $\rho_d = 50.6*10^{-12} * 1.5*10^{3}$

=> $\rho_d = 7.59*10^{-8} \ C/m$

Generally the surface charge density is mathematically represented as

$\rho_s = \frac{\rho_l }{2 \pi * r_2 }$ here $2 \pi r = (circumference \ of \ outer \ conductor )$

=> $\rho_s = \frac{7.59 *10^{-8} }{2* 3.142 * 0.003 }$

=> $\rho_s = 4.026 *10^{-6} \ C/m^2$

3 0

2 years ago

A larger car takes more force to move._

FrozenT [24]

Newton’s 2nd law
———

Newton's second law states that the acceleration of an object is directly related to the force on it, and inversely related to the mass of the object. You need more force to move or stop an object with a lot of mass (or inertia) than you need for an object with less mass. .

5 0

3 years ago

Alyse runs to her classroom which is 20 meters away. She forgets her physics book in her locker and must return 20 meters in the

notka56 [123]

The average velocity of Alyse is 0 m/s.

<h3>What is the average velocity of an object or a person?</h3>

The average velocity of a person is the average of all the velocities of moving as it moves.

Velocity on its own is the ratio of the displacement of a body and the time taken for that displacement.

Mathematically:

Average velocity = total displacement / time taken

Displacement is a vector quantity that measures the distance traveled by a moving body in a specific direction.

Hence, average velocity and velocity are both vector quantities.

Displacement takes a measure of the shortest possible distance along a straight line from the beginning of the motion until the end of the motion.

Calculating the average velocity of Alyse:

Assuming the initial direction is positive and n=backeard direction is negative:

Total displacement = 20 m + - (20 m)

Total displacement = 0 m

total time = 3 minutes

Average velocity = 0 m/ 3

Average velocity = 0 m/s

Learn more about average velocity at: brainly.com/question/4931057

#SPJ1

3 0

1 year ago