Answer:

Explanation:
The resistance of a conductor is directly proportional to its length and is inversely proportional to its cross-sectional area, this dependence is given by:

is the material's resistance, L is the legth and A is the cross-sectional area.
For the first and second coils, we have:

For the third and fourth coils, we have:

<span>A design is remodeled after analysis and tested again.</span>
Answer:
Explanation:
For entry of light into tube of unknown refractive index
sin ( 90 - 25 ) / sinr = μ , μ is the refractive index of the tube , r is angle of refraction in the medium of tube
r = 90 - C where C is critical angle between μ and body medium in which tube will be inserted.
sin ( 90 - 25 ) / sin( 90 - C) = μ
sin65 / cos C = μ
sinC = 1.33 / μ , where 1.33 is the refractive index of body liquid.
From these equations
sin65 / cos C = 1.33 / sinC
TanC = 1.33 / sin65
TanC = 1.33 / .9063
TanC = 1.4675
C= 56°
sinC = 1.33 / μ
μ = 1.33 / sinC
= 1.33 / sin56
= 1.33 / .829
μ = 1.6 Ans
The answer is 2400 centimeters
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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