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Answer:
D ≈ 8.45 m
L ≈ 100.02 m
Explanation:
Given
Q = 350 m³/s (volumetric water flow rate passing through the stretch of channel, maximum capacity of the aqueduct)
y₁ - y₂ = h = 2.00 m (the height difference from the upper to the lower channels)
x = 100.00 m (distance between the upper and the lower channels)
We assume that:
- the upper and the lower channels are at the same pressure (the atmospheric pressure).
- the velocity of water in the upper channel is zero (v₁ = 0 m/s).
- y₁ = 2.00 m (height of the upper channel)
- y₂ = 0.00 m (height of the lower channel)
- g = 9.81 m/s²
- ρ = 1000 Kg/m³ (density of water)
We apply Bernoulli's equation as follows between the point 1 (the upper channel) and the point 2 (the lower channel):
P₁ + (ρ*v₁²/2) + ρ*g*y₁ = P₂ + (ρ*v₂²/2) + ρ*g*y₂
Plugging the known values into the equation and simplifying we get
Patm + (1000 Kg/m³*(0 m/s)²/2) + (1000 Kg/m³)*(9.81 m/s²)*(2 m) = Patm + (1000 Kg/m³*v₂²/2) + (1000 Kg/m³)*(9.81 m/s²)*(0 m)
⇒ v₂ = 6.264 m/s
then we apply the formula
Q = v*A ⇒ A = Q/v ⇒ A = Q/v₂
⇒ A = (350 m³/s)/(6.264 m/s)
⇒ A = 55.873 m²
then, we get the diameter of the pipe as follows
A = π*D²/4 ⇒ D = 2*√(A/π)
⇒ D = 2*√(55.873 m²/π)
⇒ D = 8.434 m ≈ 8.45 m
Now, the length of the pipe can be obtained as follows
L² = x² + h²
⇒ L² = (100.00 m)² + (2.00 m)²
⇒ L ≈ 100.02 m
Guessing you want the average speed. We can multiple each speed by the time we spent going that speed, and them all together and then divide by the total time we spent in traffic to get the average speed. We spent a total of 7.5 minutes in traffic, so average speed = (12*1.5+0*3.5+15*2.5)/7.5 = 7.4 m/s
Answer:
Boiling Point
Explanation:
When a liquid changes to a gas is called the boiling point.
Solution:
With reference to Fig. 1
Let 'x' be the distance from the wall
Then for 
DAC:
⇒ 
Now for the BAC:
⇒ 
Now, differentiating w.r.t x:
![\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}]](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5Ctheta%20%7D%7Bdx%7D%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Btan%5E%7B-1%7D%20%5Cfrac%7Bd%20%2B%20h%7D%7Bx%7D%20-%20%20tan%5E%7B-1%7D%20%5Cfrac%7Bd%7D%7Bx%7D%5D)
For maximum angle, 
= 0
Now,
0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}][/tex]
0 = 
After solving the above eqn, we get
x = 
The observer should stand at a distance equal to x =
