<span>they are perpendicular.</span>
Answer:
9
Step-by-step explanation:
The function's y-intercept is the value for x=0.
h(0) = 0.5(0 +4)² +1
= 0.5(16) +1
= 8 +1
h(0) = 9
The function's y-intercept is 9.
ita 3 fiths x plus three eigths
Answer:
V = 20.2969 mm^3 @ t = 10
r = 1.692 mm @ t = 10
Step-by-step explanation:
The solution to the first order ordinary differential equation:
![\frac{dV}{dt} = -kA](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdt%7D%20%3D%20-kA)
Using Euler's method
![\frac{dVi}{dt} = -k *4pi*r^2_{i} = -k *4pi*(\frac {3 V_{i} }{4pi})^(2/3)\\ V_{i+1} = V'_{i} *h + V_{i} \\](https://tex.z-dn.net/?f=%5Cfrac%7BdVi%7D%7Bdt%7D%20%3D%20-k%20%2A4pi%2Ar%5E2_%7Bi%7D%20%3D%20-k%20%2A4pi%2A%28%5Cfrac%20%7B3%20V_%7Bi%7D%20%7D%7B4pi%7D%29%5E%282%2F3%29%5C%5C%20V_%7Bi%2B1%7D%20%3D%20V%27_%7Bi%7D%20%2Ah%20%2B%20V_%7Bi%7D%20%20%20%20%5C%5C)
Where initial droplet volume is:
![V(0) = \frac{4pi}{3} * r(0)^3 = \frac{4pi}{3} * 2.5^3 = 65.45 mm^3](https://tex.z-dn.net/?f=V%280%29%20%3D%20%5Cfrac%7B4pi%7D%7B3%7D%20%2A%20r%280%29%5E3%20%3D%20%20%5Cfrac%7B4pi%7D%7B3%7D%20%2A%202.5%5E3%20%3D%2065.45%20mm%5E3)
Hence, the iterative solution will be as next:
- i = 1, ti = 0, Vi = 65.45
![V'_{i} = -k *4pi*(\frac{3*65.45}{4pi})^(2/3) = -6.283\\V_{i+1} = 65.45-6.283*0.25 = 63.88](https://tex.z-dn.net/?f=V%27_%7Bi%7D%20%20%3D%20-k%20%2A4pi%2A%28%5Cfrac%7B3%2A65.45%7D%7B4pi%7D%29%5E%282%2F3%29%20%20%3D%20-6.283%5C%5CV_%7Bi%2B1%7D%20%3D%2065.45-6.283%2A0.25%20%3D%2063.88)
- i = 2, ti = 0.5, Vi = 63.88
![V'_{i} = -k *4pi*(\frac{3*63.88}{4pi})^(2/3) = -6.182\\V_{i+1} = 63.88-6.182*0.25 = 62.33](https://tex.z-dn.net/?f=V%27_%7Bi%7D%20%20%3D%20-k%20%2A4pi%2A%28%5Cfrac%7B3%2A63.88%7D%7B4pi%7D%29%5E%282%2F3%29%20%20%3D%20-6.182%5C%5CV_%7Bi%2B1%7D%20%3D%2063.88-6.182%2A0.25%20%3D%2062.33)
- i = 3, ti = 1, Vi = 62.33
![V'_{i} = -k *4pi*(\frac{3*62.33}{4pi})^(2/3) = -6.082\\V_{i+1} = 62.33-6.082*0.25 = 60.813](https://tex.z-dn.net/?f=V%27_%7Bi%7D%20%20%3D%20-k%20%2A4pi%2A%28%5Cfrac%7B3%2A62.33%7D%7B4pi%7D%29%5E%282%2F3%29%20%20%3D%20-6.082%5C%5CV_%7Bi%2B1%7D%20%3D%2062.33-6.082%2A0.25%20%3D%2060.813)
We compute the next iterations in MATLAB (see attachment)
Volume @ t = 10 is = 20.2969
The droplet radius at t=10 mins
![r(10) = (\frac{3*20.2969}{4pi})^(2/3) = 1.692 mm\\](https://tex.z-dn.net/?f=r%2810%29%20%3D%20%28%5Cfrac%7B3%2A20.2969%7D%7B4pi%7D%29%5E%282%2F3%29%20%3D%20%3Cstrong%3E1.692%20mm%3C%2Fstrong%3E%5C%5C)
The average change of droplet radius with time is:
Δr/Δt = ![\frac{r(10) - r(0)}{10-0} = \frac{1.692 - 2.5}{10} = -0.0808 mm/min](https://tex.z-dn.net/?f=%5Cfrac%7Br%2810%29%20-%20r%280%29%7D%7B10-0%7D%20%3D%20%5Cfrac%7B1.692%20-%202.5%7D%7B10%7D%20%3D%20-0.0808%20mm%2Fmin)
The value of the evaporation rate is close the value of k = 0.08 mm/min
Hence, the results are accurate and consistent!
Answer:
I would need more information but I believe its A
Step-by-step explanation: