Answer:
Fossil Combustion Reactions
Explanation:
It's more efficient (I'll edit later)
Answer:
195.168 m
Explanation:
To find the magnitude of the vector you can use the Pythagorean Theorem since you have the height and base and the vector is really just the hypotenuse
Pythagorean Theorem:
![a^2+b^2=c^2](https://tex.z-dn.net/?f=a%5E2%2Bb%5E2%3Dc%5E2)
Plug values in
![88.4^2+174^2=c^2](https://tex.z-dn.net/?f=88.4%5E2%2B174%5E2%3Dc%5E2)
Simplify
![7814.56+30276=c^2](https://tex.z-dn.net/?f=7814.56%2B30276%3Dc%5E2)
Add the two values
![38090.56=c^2](https://tex.z-dn.net/?f=38090.56%3Dc%5E2)
Take the square root of both sides
![195.168\approx195.168](https://tex.z-dn.net/?f=195.168%5Capprox195.168)
Answer:
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Integrating the velocity equation, we will see that the position equation is:
![$f(t)=\frac{\cos ^3(\omega t)-1}{3}](https://tex.z-dn.net/?f=%24f%28t%29%3D%5Cfrac%7B%5Ccos%20%5E3%28%5Comega%20t%29-1%7D%7B3%7D)
<h3>How to get the position equation of the particle?</h3>
Let the velocity of the particle is:
![$v(t)=\sin (\omega t) * \cos ^2(\omega t)](https://tex.z-dn.net/?f=%24v%28t%29%3D%5Csin%20%28%5Comega%20t%29%20%2A%20%5Ccos%20%5E2%28%5Comega%20t%29)
To get the position equation we just need to integrate the above equation:
![$f(t)=\int \sin (\omega t) * \cos ^2(\omega t) d t](https://tex.z-dn.net/?f=%24f%28t%29%3D%5Cint%20%5Csin%20%28%5Comega%20t%29%20%2A%20%5Ccos%20%5E2%28%5Comega%20t%29%20d%20t)
![$\mathrm{u}=\cos (\omega \mathrm{t})](https://tex.z-dn.net/?f=%24%5Cmathrm%7Bu%7D%3D%5Ccos%20%28%5Comega%20%5Cmathrm%7Bt%7D%29)
Then:
![$d u=-\sin (\omega t) d t](https://tex.z-dn.net/?f=%24d%20u%3D-%5Csin%20%28%5Comega%20t%29%20d%20t)
![\Rightarrow d t=-d u / \sin (\omega t)](https://tex.z-dn.net/?f=%5CRightarrow%20d%20t%3D-d%20u%20%2F%20%5Csin%20%28%5Comega%20t%29)
Replacing that in our integral we get:
![$\int \sin (\omega t) * \cos ^2(\omega t) d t$](https://tex.z-dn.net/?f=%24%5Cint%20%5Csin%20%28%5Comega%20t%29%20%2A%20%5Ccos%20%5E2%28%5Comega%20t%29%20d%20t%24)
![$-\int \frac{\sin (\omega t) * u^2 d u}{\sin (\omega t)}-\int u^2 d t=-\frac{u^3}{3}+c$](https://tex.z-dn.net/?f=%24-%5Cint%20%5Cfrac%7B%5Csin%20%28%5Comega%20t%29%20%2A%20u%5E2%20d%20u%7D%7B%5Csin%20%28%5Comega%20t%29%7D-%5Cint%20u%5E2%20d%20t%3D-%5Cfrac%7Bu%5E3%7D%7B3%7D%2Bc%24)
Where C is a constant of integration.
Now we remember that ![$u=\cos (\omega t)$](https://tex.z-dn.net/?f=%24u%3D%5Ccos%20%28%5Comega%20t%29%24)
Then we have:
![$f(t)=\frac{\cos ^3(\omega t)}{3}+C](https://tex.z-dn.net/?f=%24f%28t%29%3D%5Cfrac%7B%5Ccos%20%5E3%28%5Comega%20t%29%7D%7B3%7D%2BC)
To find the value of C, we use the fact that f(0) = 0.
![$f(t)=\frac{\cos ^3(\omega * 0)}{3}+C=\frac{1}{3}+C=0](https://tex.z-dn.net/?f=%24f%28t%29%3D%5Cfrac%7B%5Ccos%20%5E3%28%5Comega%20%2A%200%29%7D%7B3%7D%2BC%3D%5Cfrac%7B1%7D%7B3%7D%2BC%3D0)
C = -1 / 3
Then the position function is:
![$f(t)=\frac{\cos ^3(\omega t)-1}{3}](https://tex.z-dn.net/?f=%24f%28t%29%3D%5Cfrac%7B%5Ccos%20%5E3%28%5Comega%20t%29-1%7D%7B3%7D)
Integrating the velocity equation, we will see that the position equation is:
![$f(t)=\frac{\cos ^3(\omega t)-1}{3}](https://tex.z-dn.net/?f=%24f%28t%29%3D%5Cfrac%7B%5Ccos%20%5E3%28%5Comega%20t%29-1%7D%7B3%7D)
To learn more about motion equations, refer to:
brainly.com/question/19365526
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