Answer:
A title
Explanation:
Because this is middle school.
<u>This is what I think is harmful!
</u>
<u></u>1.) <u>Deadly gases</u> (Example: C<span>arbon Monoxide</span>)
2.) <u>Poison
</u>
3.) <u>Snake bites</u>
(There is to many to list :D)
Have a good Day/Night :D
Answer:
3.77 Hz
Explanation:
The beat frequency that is heard can be calculated using the following equation:

<u>Where:</u>
f₁ = 110 Hz

<u>With:</u>
<em>T₁ = 594 N </em>
<em>T₂ = 554 N </em>
Hence, the beat frequency is:
Therefore, the beat frequency that is heard when the hammer strikes the two strings simultaneously is 3.77 Hz.
I hope it helps you!
Diffusion probably be more specific
Complete Question
While helping an astronomy professor, you discover a binary star system in which the two stars are in circular orbits about the system's center of mass. From their color and brightness, you determine that each star has the same mass as our Sun. The orbital period of the pair is 29.3 days , based on the oscillation of brightness observed as one star occludes (hides) the other. From this information you are able to ascertain the distance between the stars.
Calculate the distance between the stars.
Express your answer to two significant digits and include the appropriate units.
Answer:
The distance between the stars is 
Explanation:
From this question we are told that
The orbital period is 
The mass of the stars are = Mass of sun 
For the two stars to keep on rotating on the circular orbit, the gravitational force must equal to the centripetal force and this can be mathematically represented as

Where r is the radius of the circular orbit
G is the gravitational constant 
d is the linear distance between the two stars = 2 r
This because during their oscillation around the circular orbit one usually hides the other

Now substituting values into the above relation
![\frac{Gm^2}{(2r)^2} = \frac{m[\frac{2 \pi r}{T} ]^2}{r}](https://tex.z-dn.net/?f=%5Cfrac%7BGm%5E2%7D%7B%282r%29%5E2%7D%20%3D%20%5Cfrac%7Bm%5B%5Cfrac%7B2%20%5Cpi%20r%7D%7BT%7D%20%5D%5E2%7D%7Br%7D)
![\frac{Gm^2}{(2r)^2} = \frac{m[\frac{4 \pi^2 r^2 }{T} ]}{r}](https://tex.z-dn.net/?f=%5Cfrac%7BGm%5E2%7D%7B%282r%29%5E2%7D%20%3D%20%5Cfrac%7Bm%5B%5Cfrac%7B4%20%5Cpi%5E2%20r%5E2%20%20%7D%7BT%7D%20%5D%7D%7Br%7D)
![r^3 = [\frac{GmT^2}{16 \pi^2} ]](https://tex.z-dn.net/?f=r%5E3%20%3D%20%5B%5Cfrac%7BGmT%5E2%7D%7B16%20%5Cpi%5E2%7D%20%5D)
Substituting values we have
![r = [\frac{(6.67*10^{-11})(1.99*10^{30})(2531520)^2}{16(3.142)^2} ]^{\frac{1}{3} }](https://tex.z-dn.net/?f=r%20%3D%20%5B%5Cfrac%7B%286.67%2A10%5E%7B-11%7D%29%281.99%2A10%5E%7B30%7D%29%282531520%29%5E2%7D%7B16%283.142%29%5E2%7D%20%5D%5E%7B%5Cfrac%7B1%7D%7B3%7D%20%7D)
![=\sqrt[3]{5.38529*10^{30}}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B3%5D%7B5.38529%2A10%5E%7B30%7D%7D)

Now d = 2r
=> d 
