HELPP!!! Calculating voltage drop
1 answer:
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Answer:
I)E= 40.95 GPa
II)E=5.29 GPa
Explanation:
I)
Given that
E₁ = 2.41 GPa ,V₁=1-0.55 = 0.45
E₂ = 72.5 GPa ,V₂=0.55
Longitudinal moduli given as ;
E= E₁V₁+E₂V₂
E= 2.41 x 0.45 + 72.5 x 0.55 GPa
E= 40.95 GPa
II)
E₁ = 2.41 GPa ,V₁=1-0.55 = 0.45
E₂ =230 GPa ,V₂=0.55
Transverse moduli given as:
E=5.29 GPa
Answer:
Total wight =640.7927 KN
Explanation:
Given that
do= 61 cm
L =120
t= 0.9 cm
That is why inner diameter of the pipe
di= 61 - 2 x 0.9 cm
di=59.2 cm
Water density ,ρ = 1 kg/L = 1000 kg/m³
Weight of the pipe ,wt = 2500 N/m
wt = 2500 x 120 N = 300,000 N
The wight of the water
wt ' = ρ V g
wt'=340792.47 N
That is why total wight
Total wight = wt + wt'
Total wight =300,000+ 340792.47 N
Total wight =640,792.47 N
Total wight =640.7927 KN