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:
Explanation:
A woman walks due west on the deck of a ship at 3 miyh.The ship is moving north at a speed of 22 miyh.Find the speed and direction of the woman relative to the surface of the water.
Answer:
The maximum length is 3.897×10^-5 mm
Explanation:
Extension = surface energy/elastic modulus
surface energy = 1.05 J/m^2
elastic modulus = 198 GPa = 198×10^9 Pa
Extension = 1.05/198×10^9 = 5.3×10^-12 m
Strain = stress/elastic modulus = 27×10^6/198×10^9 = 1.36×10^-4
Length = extension/strain = 5.3×10^-12/1.36×10^-4 = 3.897×10^-8 m = 3.897×10^-8 × 1000 = 3.897×10^-5 mm