Answer:

Step-by-step explanation:

Answer:
|F net| = 20.22 N
θ ≈ 19.8°
Step-by-step explanation:
F net = 15N i + 8cos(60°)N i + 8sin(60°)N j
= 15N i + 8×½N i + 8×√3/2N j
= 15N i + 4N i + 4√3N j
= 19N i + 4√3N j
|F net| = √(19²+(4√3)²) = √(361+48) = √409 ≈ 20.22N
tan(θ) = 4√3 ÷ 19 ≈ 0.36 → θ ≈ arctan(0.36) = 19.8°
A=bh
1056=32<span>×b
</span>÷32 ÷32
3<span>3=h
The height is </span>3<span>3 inches.</span>
Answer: it triples by 3 so 15,45, 135
Step-by-step explanation: this is by 3 because it goes on by 15,45,135
Answer: ![v=\sqrt[]{\frac{2K}{m} }](https://tex.z-dn.net/?f=v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2K%7D%7Bm%7D%20%7D)
Step-by-step explanation:

First, multiply by 2 to get rid of the 2 in the denominator. Remember that if you make any changes you have to make sure the equation keeps balanced, so do it on both sides as following;


Divide by m to isolate
.


To eliminate the square and isolate v, extract the square root.
![\sqrt[]{\frac{2K}{m} }=\sqrt[]{v^2}](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7B2K%7D%7Bm%7D%20%7D%3D%5Csqrt%5B%5D%7Bv%5E2%7D)
![\sqrt[]{\frac{2K}{m} }=v](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7B2K%7D%7Bm%7D%20%7D%3Dv)
let's rewrite it in a way that v is in the left side.
![v=\sqrt[]{\frac{2K}{m} }](https://tex.z-dn.net/?f=v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2K%7D%7Bm%7D%20%7D)