Answer:
The rate at which this vacuum cleaner supply energy to the room when running is 285.71 Watts.
Explanation:
power efficiency of electric motor = 70% = 0.70
The power output of the vacuum cleaner =
= 200 W
The power output of the vacuum cleaner = 
The rate at which this vacuum cleaner supply energy to the room when running is 285.71 Watts.
Answer:
it is f all of the above
Explanation:
let me know if im right
im not positive if im right but i should be right
Answer:
y ≈ 2.5
Explanation:
Given data:
bottom width is 3 m
side slope is 1:2
discharge is 10 m^3/s
slope is 0.004
manning roughness coefficient is 0.015
manning equation is written as
where R is hydraulic radius
S = bed slope
P is perimeter 
![Q = (2+2y) y) \times 1/0.015 [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} 0.004^{1/2}](https://tex.z-dn.net/?f=Q%20%3D%20%282%2B2y%29%20y%29%20%5Ctimes%201%2F0.015%20%5B%5Cfrac%7B%283%2B2y%29%20y%7D%7B%283%2B2%5Csqrt%7B5%7D%20y%29%7D%5D%5E%7B2%2F3%7D%200.004%5E%7B1%2F2%7D)
solving for y![100 =(2+2y) y) \times (1/0.015) [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} \times 0.004^{1/2}](https://tex.z-dn.net/?f=100%20%3D%282%2B2y%29%20y%29%20%5Ctimes%20%281%2F0.015%29%20%5B%5Cfrac%7B%283%2B2y%29%20y%7D%7B%283%2B2%5Csqrt%7B5%7D%20y%29%7D%5D%5E%7B2%2F3%7D%20%5Ctimes%200.004%5E%7B1%2F2%7D)
solving for y value by using iteration method ,we get
y ≈ 2.5
