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3 years ago

28. Sound can be heard around a corner because of

Physics

1 answer:

Lesechka [4]3 years ago

5 0

Answer:

Diffraction of sound wavelengths.

Explanation:

$Diffraction$ -A wave is able to bend around a corner due to the effects of diffraction. sound aves are capable of bending around corners in the same magnitude as it's wavelength making it possible to hear sounds around corners.

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I dont know how to do this at all please help

worty [1.4K]

Wow ! I understand your shock. I shook and vibrated a little
when I looked at this one too.

The reason for our shock is all the extra junk in the question,
put there just to shock and distract us.

"Neutron star", "5.5 solar masses", "condensed burned-out star".
That's all very picturesque, and it excites cosmic fantasies in
out brains when we read it, but it's just malicious decoration.
It only gets in the way, and doesn't help a bit.

The real question is:

What is the acceleration of gravity 2000 m from
the center of a mass of 1.1 x 10³¹ kg ?

Acceleration of gravity is

   G · M / R²

= (6.67 x 10⁻¹¹ N·m²/kg²) · (1.1 x 10³¹ kg) / (2000 m)²

= (6.67 x 10⁻¹¹ · 1.1 x 10³¹ / 4 x 10⁶) (N) · m² · kg / kg² · m²

=   1.83 x 10¹⁴ (kg · m / s²) · m² · kg / kg² · m²

= 1.83 x 10¹⁴    m / s²

That's about 1.87 x 10¹³ times the acceleration of gravity on
Earth's surface.

In other words, if I were standing on the surface of that neutron star,
I would weigh 1.82 x 10¹² tons, give or take.

3 0

2 years ago

A fluid in an aquifer is 23.6 m above a reference datum, the fluid pressure (in gage pressure) is 4390 n/m2 and the flow velocit

Phoenix [80]

As per Bernuolli's Theorem total energy per unit mass is given as

$\frac{P}{\rho} + \frac{1}{2}v^2 + gH = E$

now from above equation

$P = 4390 N/m^2$

$\rho = 0.999 * 10^3$

$v = 7.22 * 10^{-4} m/s$

$H = 23.6 m$

now by above equation

$\frac{4390}{0.999*10^3} + \frac{1}{2}*(7.22*10^{-4})^2 + 9.8*23.6 = E$

$E = 235.7 J/kg$

Part B)

Now energy per unit weight

$U = \frac{E}{g}$

$U = \frac{235.7}{9.8}$

$U = 24 m$

7 0

3 years ago

Someone help me in this question.

tekilochka [14]

Gradpoint ? And the answer is water.

4 0

3 years ago

Read 2 more answers

in the system to the right, the pulleys are frictionless and the system hangs in equilibrium. Determine the values of each of th

irga5000 [103]

Look up pulleys problem through Khan academy and a video should pop up with a problem similar and you should be able to walk through it .

4 0

2 years ago

Wind instruments like trumpets and saxophones work on the same principle as the "tube closed on one end" that we examined in our

Gnom [1K]

Answer:

f = v / 4L

the frequency of the instruments is reduced by the decrease in the speed of the wave with the temperature.

Explanation:

In wind instruments the wave speed must meet

v = λ f

λ = v / f

from v is the speed of sound that depends on the temperature

v = v₀ $\sqrt{1+ \frac{T [C]}{273} }$

where I saw the speed of sound at 0ºC v₀ = 331 m/s the temperature is in degrees centigrade, we can take the degrees Fahrenheit to centigrade with the relation

(F -32) 5/9 = C

76ºF = 24.4ºC

45ºF = 7.2ºC

With this relationship we can see that the speed of sound is significantly reduced when leaving the house to the outside

at T₁ = 24ºC v₁ = 342.9 m / s

at T₂ = 7ºC v₂ = 339.7 m / s

To satisfy this speed the wavelength of the sound must be reduced, so the resonant frequencies change

λ / 4 = L

λ= 4L

v / f = 4L

f = v / 4L

Therefore, the frequency of the instruments is reduced by the decrease in the speed of the wave with the temperature.

3 0

3 years ago