Answer:
a) 199.04 ohms
b) attached in image
c) -0.696dB
Explanation:
We are given:
Fc = 8Khz = 8000hz

a)Using the formula:



R = 199.04 ohms
b) diagram is attached
c) 

At F = 20KHz and Fc= 8KHz we have:


=0.923
|H(F)| in dB = 20log |H(F)|
=20log0.923
= -0.696dB
Answer:
a) 42.08 ft/sec
b) 3366.33 ft³/sec
c) 0.235
d) 18.225 ft
e) 3.80 ft
Explanation:
Given:
b = 80ft
y1 = 1 ft
y2 = 10ft
a) Let's take the formula:

1 + 8f² = (20+1)²
= 8f² = 440
f² = 55
f = 7.416
For velocity of the faster moving flow, we have :
V1 = 42.08 ft/sec
b) the flow rate will be calculated as
Q = VA
VA = V1 * b *y1
= 42.08 * 80 * 1
= 3366.66 ft³/sec
c) The Froude number of the sub-critical flow.
V2.A2 = 3366.66
Where A2 = 80ft * 10ft
Solving for V2, we have:
= 4.208 ft/sec
Froude number, F2 =
F2 = 0.235
d)
= 18.225ft
e) for critical depth, we use :
= 3.80 ft
Answer:
Decreased risk of structure failure