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:
Explanation:
The temperature can be defined as the measurement of the intensity of the heat present in the object. Fahrenheit, kelvin and centigrade are the common scale used for measuring Temperature.
Given:
T1=170C
To convert to Kelvin
= 17+273 =290K
T1 = 290K
Pressure (P)= 95KPa
Specific heat ratio = CP/CV= K
WhereK=1.005/0.718
K = 1.4
The final temperature can be calculated using the formula below.
T2 = CP/CV × T1
=. K × T1
T2 = 1.4 × 290
Answer:it forms a molten mold that makes it hard to be able to smash something into it then make something like a key
Explanation:
Answer:
The pressure drop is 269.7N/m^2
Explanation:
∆P = ∆h × rho × g
∆h = 3.2cm = 3.2/100 = 0.032m, rho = 860kg/m^3, g = 9.8m/s^2
∆P = 0.032×860×9.8 = 269.7N/m^2
