Answer:
a) 6498.84 kW
b) 0.51
c) 0.379
Explanation:
See the attached picture below for the solution
Answer:
H_w = 2.129 m
Explanation:
given,
Width of the weir, B = 1.2 m
Depth of the upstream weir, y = 2.5 m
Discharge, Q = 0.5 m³/s
Weir coefficient, C_w = 1.84 m
Now, calculating the water head over the weir
![Q = C_w BH^{3/2}](https://tex.z-dn.net/?f=Q%20%3D%20C_w%20BH%5E%7B3%2F2%7D)
![H = (\dfrac{Q}{C_wB})^{2/3}](https://tex.z-dn.net/?f=H%20%3D%20%28%5Cdfrac%7BQ%7D%7BC_wB%7D%29%5E%7B2%2F3%7D)
![H = (\dfrac{0.5}{1.84\times 1.2})^{2/3}](https://tex.z-dn.net/?f=H%20%3D%20%28%5Cdfrac%7B0.5%7D%7B1.84%5Ctimes%201.2%7D%29%5E%7B2%2F3%7D)
![H = 0.371\ m](https://tex.z-dn.net/?f=H%20%3D%200.371%5C%20m)
now, level of weir on the channel
H_w = y - H
H_w = 2.5 - 0.371
H_w = 2.129 m
Height at which weir should place is equal to 2.129 m.
A sphere is charged with electrons to −9 × 10−6 C. The value given is the total charge of all the electrons present in the sphere. To calculate the number of electrons in the sphere, we divide the the total charge with the charge of one electron.
N = 9 × 10−6 C / 1.6 × 10−19 C
N = 5.6 x 10^13
The answer is A because the paper does not change its chemical properties only changes the way it looks.
Answer:
option B
Explanation:
given,
Satellite B has an orbital radius nine times that of satellite A.
R' = 9 R
now, orbital velocity of the satellite A
........(1)
now, orbital velocity of satellite B
from equation 1
hence, the correct answer is option B