The elevation in reservoir at the rate of flow using is 03m/s is 114m.
The Reynolds range is the ratio of inertial forces to viscous forces. The Reynolds variety is a dimensionless variety used to categorize the fluids structures in which the impact of viscosity is crucial in controlling the velocities or the flow sample of a fluid.
The reason of the Reynolds number is to get a few experience of the relationship in fluid glide between inertial forces (this is those that maintain going by using Newton's first law – an item in motion stays in movement) and viscous forces, this is people who cause the fluid to come back to a forestall because of the viscosity of the fluid.
calculation,
Let L = 100 m pipe
L1 = 150 m pipe
H f = friction losses
Using Reynolds number, relative roughness, friction co- effiicients and friction losses
Substitute the value in equation
Z = 110= 0.48= 3.54
Z = 114m
Therefore water surface elevation at reservoir is 114 meter.
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Answer:
a) U = 735 J
, b) U = 125.7 J
, c) U = 0 J
Explanation:
The gravitational power energy is
U = mg y - mg y₀
The last value is a constant, for simplicity we can make it zero, if the lowest point is at the origin of the coordinate system, which in this case we will place in the lowest part
a) Rope is horizontal
The height in this case is the same length of the rope
y = 2.10 m
w = mg = 350 N
U = 350 2.10
U = 735 J
b) when the angle is 34º
y = L - L cos 34
y = L (1- cos34)
y = 2.10 (1- cos 34)
y = 0.359 m
U = 350 0.359
U = 125.7 J
c) in this case this point coincides with the reference system
y = 0
U = 0 J
Answer: J.J Thomson
Explanation: J. J. Thomson, who discovered the electron in 1897, proposed the plum pudding model of the atom in 1904 before the discovery of the atomic nucleus in order to include the electron in the atomic model.
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Answer:
220 A
Explanation:
The magnetic force on the floating rod due to the rod held close to the ground is F = BI₁L where B = magnetic field due to rod held close the ground = μ₀I₂/2πd where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, I₂ = current in rod close to ground and d = distance between both rods = 11 mm = 0.011 m. Also, I₁ = current in floating rod and L = length of rod = 1.1 m.
So, F = BI₁L
F = (μ₀I₂/2πd)I₁L
F = μ₀I₁I₂L/2πd
Given that the current in the rods are the same, I₁ = I₂ = I
So,
F = μ₀I²L/2πd
Now, the magnetic force on the floating rod equals its weight , W = mg where m = mass of rod = 0.10kg and g = acceleration due to gravity = 9.8 m/s²
So, F = W
μ₀I²L/2πd = mg
making I subject of the formula, we have
I² = 2πdmg/μ₀L
I = √(2πdmg/μ₀L)
substituting the values of the variables into the equation, we have
I = √(2π × 0.011 m × 0.1 kg × 9.8 m/s²/[4π × 10⁻⁷ H/m × 1.1 m])
I = √(0.01078 kgm²/s²/[2 × 10⁻⁷ H/m × 1.1 m])
I = √(0.01078 kgm²/s²/[2.2 × 10⁻⁷ H])
I = √(0.0049 × 10⁷kgm²/s²H)
I = √(0.049 × 10⁶kgm²/s²H)
I = 0.22 × 10³ A
I = 220 A
