Ask question

Ask question

All categories

3 years ago

5

A forest fire makes a loud roaring sound. The explosive processes that release energy from the Sun occur at a much higher temper

ature. Why don’t you hear a roaring sound from the sun?

2 answers:

snow_tiger [21]3 years ago

8 0

Sound needs a material medium to propagate through. There is no medium in space, so sound from the sun can't GO anywhere.
The other interesting question is how heat and light from the sun DO get here.

KengaRu [80]3 years ago

5 0

Becuase for sound to happen you need air, And there is no oxygen is space.

You might be interested in

A glass ball weighs 188 g in air, 116 g when immersed in water, and 125 g when immersed in turpentine. Calculate the density of

Amanda [17]

<span>volume of a sphere = 4/3pi*r^3, so if you double the diameter (same as doubling the radius) it's 2^3 times bigger,</span>

5 0

3 years ago

A uniform electric field exists everywhere in the x, y plane. This electric field has a magnitude of 4500 N/C and is directed in

Anton [14]

Answer with Explanation:

We are given that

$E=4500 N/C$

$q=-5.5\times 10^{-9} C$

a.x=-0.2 m

$E'=\frac{Kq}{r^2}=\frac{9\times 10^9\times 5.5\times 10^{-9}}{(0.2)^2}=1237.5 N$

Where $k=9\times 10^9$

Net electric field=E+E'=4500+1237.5=5737.5 N/C

b.x=0.20 m

Net electric field=E-E'=4500-1237.5=3262.5 N/C

c.y=0.20 m

E=1237.5 N/C

Net electric field= $\sqrt{E^2+E'^2}=\sqrt{(4500)^2+(1237.5)^2}=4667.05 N/C$

7 0

3 years ago

A concert loudspeaker suspended high off the ground emits 33.0 W of sound power. A small microphone with a 0.600 cm2 area is 52.

Novay_Z [31]

Answer:

The sound intensity at the position of the microphone is $9.71\times10^{-4} W/m^{2}$

Explanation

Sound intensity is given by the formula

$I=\frac{P}{A}$

Where $I$ is the sound intensity, $P$ is the power and $A$ is the area.

Since the loudspeaker radiates sound in all directions, we have a spherical sound wave where the radius r is the distance of the microphone from the speaker.

∴ $A$ is given by $4\pi r^{2}$ where $r$ is the radius

From the question, $P$ = 33.0W, $r$ = 52.0m

$I=\frac{P}{A} = \frac{P}{4\pi r^{2} }$

$I = \frac{33.0}{4\pi \times (52.0)^{2} }$

∴ $I = 9.71\times10^{-4} W/m^{2}$

Hence, the sound intensity at the position of the microphone is 9.71 × 10⁻⁴ W/m²

7 0

3 years ago

A scientific law tells us what nature will do under certain conditions.<br> True<br> False

sergij07 [2.7K]

Hello There!

This Is A "True" Statement.

<em>A scientific law tells us what nature will do under certain conditions.</em>

6 0

3 years ago

The intensity of the sound from a certain source is measured at two points along a line from the source. The points are separate

Artemon [7]

Answer:

The source is at a distance of 4.56 m from the first point.

Solution:

As per the question:

Separation distance between the points, d = 11.0 m

Sound level at the first point, L = 66.40 dB

Sound level at the second point, L'= 55.74 dB

Now,

$L = 10log_{10}\frac{I}{I_{o}}$

$I = I_{o}10^{\frac{L}{10}} = I_{o}10^{0.1L} = 10^{- 12}\times 10^{0.1\times 66.40} = 10^{- 5.36}$

$L' = 10log_{10}\frac{I'}{I_{o}}$

$I' = I_{o}10^{\frac{L'}{10}} = 10^{- 12}\times 10^{0.1\times 55.74} = 10^{- 6.426}$

where

$I_{o} = 10^{- 12} W/m^{2}$

I = Intensity of sound

Now,

$I = \frac{P}{4\pi R^{2}}$

Similarly,

$I' = \frac{P}{4\pi (R + 11.0)^{2}}$

Now,

$\frac{I}{I'} = \frac{(R + 11.0)^{2}}{R^{2}}$

$\frac{10^{- 5.36}}{10^{- 6.426}} = \frac{(R + 11.0)^{2}}{R^{2}}$

$R ^{2} + 22R + 121 = 11.64R^{2}}$

$10.64R ^{2} - 22R - 121 = 0$

Solving the above quadratic eqn, we get:

R = 4.56 m

8 0

3 years ago