Known :
D = 12 in = 1 ft
L = 850 ft
Q = 5.6 cfs
hA = 750 ft
hB = 765 ft
PA = 85 psi = 12240 lb/ft²
Solution :
A = πD² / 4 = π(1²) / 4
A = 0.785 ft²
<u>Velocity of water :</u>
U = Q / A = 5.6 / 0.785
U = 7.134 ft/s
<u>Friction loss due to pipe length :</u>
Re = UD / v = (7.134)(1) / (0.511 × 10^(-5))
Re = 1.4 × 10⁶
(From Moody Chart, We Get f = 0.015)
hf = f(L / d)(U² / 2g) = 0.015(850 / 1)((7.134²) / 2(32.2))
hf = 10 ft
PA + γhA = PB + γhB + γhf
PB = PA + γ(hA - hB - hf)
PB = 12240 + (62.4)(750 - 765 - 10)
PB = 10680 lb/ft²
PB = 74.167 psi
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Explanation:i hop this helps
Answer:
the correct answer is not in the options
the answer is supposed to be 10m/s² from the formula F=ma
Answer:
It will not experience fracture when it is exposed to a stress of 1030 MPa.
Explanation:
Given
Klc = 54.8 MPa √m
a = 0.5 mm = 0.5*10⁻³m
Y = 1.0
This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1030 MPa, given the values of <em>KIc</em>, <em>Y</em>, and the largest value of <em>a</em> in the material. This requires that we solve for <em>σc</em> from the following equation:
<em>σc = KIc / (Y*√(π*a))</em>
Thus
σc = 54.8 MPa √m / (1.0*√(π*0.5*10⁻³m))
⇒ σc = 1382.67 MPa > 1030 MPa
Therefore, the fracture will not occur because this specimen can handle a stress of 1382.67 MPa before experience fracture.
Answer: Either your computer is malfunctioning or it is glitched and still thinks you only have 2. Also good job
Explanation: Leave a brainliest it helps
