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Kaylis [27]
3 years ago
8

Audible limit of the sound wave

Physics
2 answers:
Sergio039 [100]3 years ago
6 0

Answer:

20 hz - 20khz is the audible sound wave for humans

o-na [289]3 years ago
3 0

Almost 18Hz is an audible limit of the sound wave.

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What is the order of magnitude of the distance of Sun to nearest star in meters?
neonofarm [45]

Answer:

Approximating the Milky Way as a disk and using the density in the solar neighborhood, there are about 100 billion stars in the Milky Way.

Explanation:

Since we are making an order of magnitude estimate, we will make a series of simplifying assumptions to get an answer that is roughly right.

Let's model the Milky Way galaxy as a disk.

The volume of a disk is:

V

=

π

⋅

r

2

⋅

h

Plugging in our numbers (and assuming that

π

≈

3

)

V

=

π

⋅

(

10

21

m

)

2

⋅

(

10

19

m

)

V

=

3

×

10

61

m

3

Is the approximate volume of the Milky Way.

Now, all we need to do is find how many stars per cubic meter (

ρ

) are in the Milky Way and we can find the total number of stars.

Let's look at the neighborhood around the Sun. We know that in a sphere with a radius of

4

×

10

16

m there is exactly one star (the Sun), after that you hit other stars. We can use that to estimate a rough density for the Milky Way.

ρ

=

n

V

Using the volume of a sphere

V

=

4

3

π

r

3

ρ

=

1

4

3

π

(

4

×

10

16

m

)

3

ρ

=

1

256

10

−

48

stars /

m

3

Going back to the density equation:

ρ

=

n

V

n

=

ρ

V

Plugging in the density of the solar neighborhood and the volume of the Milky Way:

n

=

(

1

256

10

−

48

m

−

3

)

⋅

(

3

×

10

61

m

3

)

n

=

3

256

10

13

n

=

1

×

10

11

stars (or 100 billion stars)

Is this reasonable? Other estimates say that there are are 100-400 billion stars in the Milky Way. This is exactly what we found.

4 0
2 years ago
Read 2 more answers
I really need Brainliest i am desprate
kondor19780726 [428]

Hy tikki! I've asked some questions, so of you find the questions as easy, then answer it. I'll surely mark you as brainliest :)

6 0
3 years ago
What instrument is used to expand burr holes?
k0ka [10]
Craniotomes are used
4 0
3 years ago
A hot-air balloon is accelerating upward under the influence of two forces, its weight and the buoyant force. for simplicity, co
Elenna [48]
Let V = the volume of the balloon
Force of gravity = V * ?hot * g downward
Buoyant force = V * ?cool * g upward
Net upward force F = V * ?cool * g - V * ?hot * g

F = V g (?cool - ?hot)

Mass of the balloon m = V ?hot

a = F/m = V g (?cool - ?hot)/(V ?hot)

a = g(?cool/?hot - 1)

a = 9.8(1.29/0.93 - 1)

a = 3.79 m/s^2

<span>Answer is 3.79 m/s^2</span>
4 0
3 years ago
A capacitor with an initial potential difference of 185 V is discharged through a resistor when a switch between them is closed
GrogVix [38]

Answer:

  • a. \tau =  2.1161 s
  • b. V(18.8 \ s) = 0.0256 \ V

Explanation:

<h3>a.</h3>

The equation for the voltage V of  discharging capacitor in an RC circuit at time t is:

V(t) = V_0 e^{(- \frac{t}{\tau}) }

where V_0 is the initial voltage, and \tau is the time constant.

For our problem, we know

V_0 = 185 \ V

and

V(10 \ s) = V_0 e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V

So

185 \ V \ e^{(- \frac{10 \ s}{\tau}) } = 1.64 \ V

e^{(- \frac{10 \ s}{\tau}) } = \frac{1.64 \ V}{ 185 \ V }

ln (e^{(- \frac{10 \ s}{\tau}) } ) = ln (\frac{1.64 \ V}{ 185 \ V })

- \frac{10 \ s}{\tau}  = ln (\frac{1.64 \ V}{ 185 \ V })

\tau =  \frac{-10 \s}{ln (\frac{1.64 \ V}{ 185 \ V }) }

This gives us

\tau =  2.1161 s

and this is the time constant.

<h3>b.</h3>

At t = 18.8 s we got:

V(18.8 \ s) = 185 \ V  \ e^{(- \frac{18.8 \ s}{2.1161 s}) }

V(18.8 \ s) = 185 \ V \ e^{(- \frac{18.8 \ s}{2.1161 s}) }

V(18.8 \ s) = 0.0256 \ V

4 0
3 years ago
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