At a point on the streamline, Bernoulli's equation is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = density of air, 0.075 lb/ft³ (standard conditions)
g = 32 ft/s²
Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s
Point 2 (stagnation):
At the stagnation point, the velocity is zero.
The density remains constant.
Let p₂ = pressure at the stagnation point.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
= 314.37 lb/ft²
= 314.37/144 = 2.18 lb/in²
Answer: 2.2 psi
The heavy stone would produce waves with a higher amplitude, rather than the smaller stone, because since the stone is heaver its going to have a grater impact and displace more water to create a bigger wave.
Answer:
the 70kg man
Explanation:
because he has more weight and is moving faster
Answer:
Apparent depth (Da) = 60.15 cm (Approx)
Explanation:
Given:
Distance from fish (D) = 80 cm
Find:
Apparent depth (Da)
Computation:
We know that,
Refractive index of water (n2) = 1.33
So,
Apparent depth (Da) = D(n1/n2)
Apparent depth (Da) = 80 (1/1.33)
Apparent depth (Da) = 60.15 cm (Approx)
Answer:
height of the water rise in tank is 10ft
Explanation:
Apply the bernoulli's equation between the reservoir surface (1) and siphon exit (2)
-------(1)
substitute 
0ft/s for V₁, 20ft for (z₁ - z₂) and 32.2ft/s² for g in eqn (1)
Applying bernoulli's equation between tank surface (3) and orifice exit (4)
substitute
0ft/s for V₃, h for z₃, 0ft for z₄, 32,2ft/s² for g
At equillibrium Fow rate at point 2 is equal to flow rate at point 4
Q₂ = Q₄
A₂V₂ = A₃V₃
The diameter of the orifice and the siphon are equal , hence there area should be the same
substitute A₂ for A₃
for V₂
for V₄
A₂V₂ = A₃V₃
Therefore ,height of the water rise in tank is 10ft
