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 :
![y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3](https://tex.z-dn.net/?f=%20y_c%20%3D%20%5B%5Cfrac%7B%28%5Cfrac%7B3366.66%7D%7B80%7D%29%5E2%7D%7B32.2%7D%5D%5E1%5E%2F%5E3%20)
= 3.80 ft
Answer:
True :)
Explanation:
You can recycle it! Tire recycling is the most practical and environment-friendly way of disposing of old and worn-out tires. Due to their inherent durability, large volume and environment and health risks, tires are one of the most problematic sources of solid wastes.
Hope it helped have a nice day! :)
Answer: y = x * 1dollars - 30dollars
Explanation:
Giving that the delivery cost in dollar is potent for all x > = 50 pounds of wght
Y = (x - 50)*1 dollar + c ...equ 1
Y = delivery cost equation in dollars
x = weigt of baggage for delivery
c = 20dollars = down payment for the first 50 pound weight of baggage
Equ 1 becomes
Y = (x)dollars - 50 dollars + 20 dollars
Y = (x) dollars -30 dollars
An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s2 . The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s2
Find the total time required for the police car to over take the automobile.
Answer:
15.02 sec
Explanation:
The total time required for the police car to overtake the automobile is related to the distance covered by both cars which is equal from instant point of abreast.
So; we can say :
By using the second equation of motion to find the distance S;
where ;
u = 0
Recall that:
= 46.68 - 7.85 t -2.505 t² = 0
Solving by using quadratic equation;
t = -6.16 OR t = 3.02
Since we can only take consideration of the value with a positive integer only; then t = 3.02 secs
From the question; The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit;
Therefore ; the total time required for the police car to over take the automobile = 12 s + 3.02 s
Total time required for the police car to over take the automobile = 15.02 sec
