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iogann1982 [59]

2 years ago

12

A camper stands in a valley between two parallel cliff walls. He claps his hands and notices that the echo from the nearby wall

returns 0.9 s later while the echo from the farther wall returns 1.10 s later. If the air temperature is -10 C, how wide is the valley?

1 answer:

Stells [14]2 years ago

6 0

5,658 ft is your answer I believe

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Conductors have ___<br> resistance.

vazorg [7]

Answer:

little/no

Explanation:

Conductors are materials, which conduct electricity and/or heat. That means, that their resistance to such energy is so little, that an electric current is able to pass through.

5 0

3 years ago

Your boss asks you to design a room that can be as soundproof as possible and provides you with three samples of material. The o

Ierofanga [76]

<span>C. Sample C would be best, because the percentage of the energy
in an incident wave that remains in a reflected wave from this material
is the smallest.

The coefficient of absorption is the percentage of incident sound
that's absorbed. So the highest coefficient of absorption results in
the smallest </span><span>percentage of the energy in an incident wave that remains.
That's what you want. </span>

6 0

3 years ago

Which of the following are true statements? A. Like charges repel B. Unlike charges repel C. Unlike charges attract or D. Charge

Brums [2.3K]

Answer: The correct answers are (A) and (C).

Explanation:

The expression from electrostatic force is as follows;

$F=\frac{kq_{1}q_{1}}{r^{2} }$

Here, F is the electrostatic force, k is constant, r is the distance between the charges and $q_{1},q_{1}$ are the charges.

The electrostatic force follows inverse square law. It is inversely proportional to the square of the distance between the charges. It is directly proportional to the product of the charges.

Like charges repel each other. There is a force of electrostatic repulsion between the like charges. Unlike charges attract each other. There is a force of electrostatic attraction between unlike charges.

The charges are induced on the neutral object when it is placed nearby the charged object without actually touching it.

Therefore, the true statements from the given options are as follows;

Like charges repel.

Unlike charges attract.

7 0

3 years ago

A railroad train is traveling at a speed of 26.0 m/s in still air. The frequency of the note emitted by the locomotive whistle i

olga55 [171]

Answer: 0.757m; 0.881m; 432.70Hz; 371.89Hz

Explanation:

Give the following :

Velocity of train (Vt) = 26m/s

Frequency of sound (Fs) = 420Hz

Speed of sound (Vs) = 344m/s

1) wavelength = (Vs - Vt) / Fs

Wavelength = (344 - 26) / 420 = 318/420 = 0.757m

11) Wavelength = (Vs + Vt) / Fs

Wavelength = (344 + 26) / 420 = 370/420 = 0.881m

111) According to the doppler effect :

Fl = [(V + Vl) / (V + Vs)] * fs

Fl = frequency of listener ; fs = frequency of sound source ; V = speed of sound ; Vl = Velocity of listener ; Vs = speed of sound source

Vs = - ve (train moving towards listener)

Fl = [(V + Vl) / (V + Vs)] * fs

Fl = [(344 + 0) / (344 - 26)] * 400

Fl = (344 / 318) * 400 = 432.70Hz

1V) Vs = + ve (train moving away listener)

Fl = [(V + Vl) / (V + Vs)] * fs

Fl = [(344 + 0) / (344 + 26)] * 400

Fl = (344 / 370) * 400 = 371.89Hz

6 0

3 years ago

Enrico Fermi (1901–1954) was a famous physicist who liked to pose what are now known as Fermi problems, in which several assumpt

Katarina [22]

Answer:

Explanation:

(a)

Since the earth is assumed to be a sphere.

Volume of atmosphere = volume of (earth +atm osphere) — volume of earth

$= \frac{4}{3}\pi(6400+ 50)^3 - \frac{4}{3}\pi (6400)
^3\\\\= \frac{4}{3}\pi(6192125000) km’^3\\= 2.6\times 10^{19} m^3$

Hence the volume of atmosphere is $2.6\times 10^{19} m^3$

(b)

Write the ideal gas equation as foll ows:

$PV = nRT\\\\n\frac{0.20atm\times 2.6\times10^{19} m^3}{0.08206L\, atm/mok\, K \times (15+273+15)K}\times \frac{1L}{10^{-3}m^3}\\\\= 2.20\times 10^{20} moles$

$no.\, of\, molecules = 2.20\times 10^{20} moles \times \frac{6.022\times10^{23}\,molecules}{1mole}= 13.3\times10^{43} molecules
$

Hence the required molecules is $13.3\times10^{43} molecules
$

(c)

Write the ideal gas equation as follows:

$PV =nRT
\\\\n=\frac{1.0 atm \times 0.5L
}{0.08206 L\, atm/mol\,K \times (37 +273.1 5)K} = 0.0196 moles$

$no.\, of\, molecules = 0.0196 moles \times\frac{6.022\times10^{23} molecules}
{Imole}= 1.2\times 10^{23} molecules$

Hence the required molecules in Caesar breath is $1.2\times 10^{23} molecules$

(d)

Volume fraction in Caesar last breath is as follows:

$Fraction,\, X =\frac{12\times 10 molecules}{13.3\times 10^{43} \,molecules}= 9.0\times 10\, molecule/air\, molecule}$

(e)

Since the volume capacity of the human body is 500 mL.

$Volume\, of\, Caesar\, nreath\, inhale\, is =\frac{ 12\times 10^{22}\, molecules}{breath}\times \frac{9.0\times10^{-23} molecule}{air\, molecule}\\\\= 1.08 molecule/breath$

5 0

3 years ago