The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.
<h3>How to determine the friction factor</h3>
Using the formula
μ = viscosity = 0. 06 Pas
d = diameter = 120mm = 0. 12m
V = velocity = 1m/s and 3m/s
ρ = density = 0.9
a. Velocity = 1m/s
friction factor = 0. 52 × 
friction factor = 0. 52 × 
friction factor = 0. 52 × 0. 55
friction factor 
b. When V = 3mls
Friction factor = 0. 52 × 
Friction factor = 0. 52 × 
Friction factor = 0. 52 × 0. 185
Friction factor 
Loss When V = 1m/s
Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter
Head loss = 0. 289 × 
× 
× 
Head loss = 1. 80 × 10^8
Head loss When V = 3m/s
Head loss = 
× 
× 
×
Head loss = 5. 3× 10^8
Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.
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Answer:
A. Splitting water into hydrogen and oxygen.
Explanation:
A chemical change produces a new substance. ... A physical change does not produce a new substance. Changes in state or phase (melting, freezing, vaporization, condensation, sublimation) are physical changes.
Answer:
Explanation:
Where E is the magnitude of electric field...
k is called Columb's Constant. It has a value of 8.99 x 109 N m2/C2.
Qs is the magnitude of the source charge...
and r is the magnitude of distance between source and target...
(When electron comes to rest Δt the magnitude of Electric field E become zero momentarily but later achieves the maximum value...)
Answer:
y = 2.74 m
Explanation:
The linear thermal expansion processes are described by the expression
ΔL = α L ΔT
Where α the thermal dilation constant for concrete is 12 10⁻⁶ºC⁻¹, ΔL is the length variation and ΔT the temperature variation in this case 20ªc
If the bridge is 250 m long and is covered by two sections each of them must be L = 125 m, let's calculate the variation in length
ΔL = 12 10⁻⁶ 125 20
ΔL = 3.0 10⁻² m
Let's use trigonometry to find the height
The hypotenuse Lf = 125 + 0.03 = 125.03 m
Adjacent leg L₀ = 125 m
cos θ = L₀ / Lf
θ = cos⁻¹ (L₀ / Lf)
θ = cos⁻¹ (125 / 125.03)
θ = 1,255º
We calculate the height
tan 1,255 = y / x
y = x tan 1,255
y = 125 tan 1,255
y = 2.74 m
