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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Explanation:
d) Magnetic force is the power that pulls materials together (magnet e. g iron)
an example :how magnet can pick up a coin.
e) frictional force produces when two surfaces are in contact with each other.
effects of friction : I) it produces heat
II) it causes loss in power.
Since the fourmula is mass/volume=density multiplying the density and volume yields the mass which is <span>0.80522 grams.</span>
Explanation:
a) The rope obeys Hooke's law, so:
F = k Δx
The elastic energy in the rope is:
EE = ½ k Δx²
Or, in terms of F:
EE = ½ F Δx
Use trigonometry to find the stretched length.
cos 20° = 35 / x
x = 37.25
So the displacement is:
Δx = 37.25 − 24
Δx = 13.25
The elastic energy per rope is:
EE = ½ (3.7×10⁴ N) (13.25 m)
EE = 245,000 J
There's two ropes, so the total energy is:
2EE = 490,000 J
Rounded to one significant figure, the elastic energy is 5×10⁵ J.
b) The elastic energy in the ropes is converted to gravitational energy.
EE = PE = mgh
5×10⁵ J = (1.2×10³ kg) (9.8 m/s²) h
h = 42 m
Rounded to one significant figure, the height is 40 m. So the claim is not justified.
Answer:
P₀ = 5.76 x 10⁻² MW
Explanation:
given,
efficiency of the power plant = 32%
Power produced by the nuclear fission = 0.18 MW
the power plant output = ?
using formula of efficiency
where P is the power produced in the power plant
P₀ is the power output of the power plant
P₀ = 0.18 x 0.32
P₀ = 0.0576 MW
P₀ = 5.76 x 10⁻² MW
Power plant output is equal to P₀ = 5.76 x 10⁻² MW
