Answer:
4Al + 3Mno2 --> 2Al2o3 + 3Mn
Explanation:
Al = 1 x 4 = 4
Mn = 1 x 3 = 3
O = 2 x 3 = 6
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Al = 2 x 2 = 4
Mn = 1 x 3 = 3
O = 2 x 3 = 6
Answer:
p = 15 newts
Explanation:
p is momentum
also there's a calculator to answer these questions.
Answer:
The distance is
Explanation:
From the question we are told that
The smallest shift is ![d = 0.2 \ grid \ units](https://tex.z-dn.net/?f=d%20%3D%200.2%20%5C%20grid%20%5C%20units)
Generally a grid unit is
of an arcsec
This implies that 0.2 grid unit is ![k = \frac{0.2}{10} = 0.02 \ arc sec](https://tex.z-dn.net/?f=k%20%3D%20%20%5Cfrac%7B0.2%7D%7B10%7D%20%3D%200.02%20%20%5C%20%20arc%20sec)
The maximum distance at which a star can be located and still have a measurable parallax is mathematically represented as
![d = \frac{1}{k}](https://tex.z-dn.net/?f=d%20%3D%20%20%5Cfrac%7B1%7D%7Bk%7D)
substituting values
![d = \frac{1}{0.02}](https://tex.z-dn.net/?f=d%20%3D%20%20%5Cfrac%7B1%7D%7B0.02%7D)
![d = 50 \ parsec](https://tex.z-dn.net/?f=d%20%3D%2050%20%5C%20parsec)
Note ![1 \ parsec \ \to 3.26 \ light \ year \ \to 3.086*10^{13} \ km](https://tex.z-dn.net/?f=1%20%5C%20parsec%20%20%5C%20%5Cto%203.26%20%5C%20light%20%5C%20year%20%5C%20%5Cto%203.086%2A10%5E%7B13%7D%20%5C%20km)
So
Answer:
I hope this helps.
Explanation:
It's important to know the location of an active fault in order to determine the magnitude of the expected earthquake. There is a chance than an inactive fault can become active again. It's important that we take the locations into account in order to be prepared and ready for if it occurs.
I believe it's the the third one. :)
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