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avanturin [10]
2 years ago
14

Which of the following is a possible consequence of poor measurement in construction as stated in the segment?

Engineering
1 answer:
Kamila [148]2 years ago
6 0

Answer:

Decreased risk of structure failure

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PLEASE HELP!!! ILL GIVE BRANLIEST *EXTRA POINTS* dont skip :((
Oksi-84 [34.3K]
The first one is the answer
8 0
3 years ago
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Brainliest need help
insens350 [35]

Answer:

answer c

Explanation:

4 0
2 years ago
Seawater containing 3.50 wt% salt passes through a series of 11 evaporators. Roughly equal quantities of water are vaporized in
statuscvo [17]

Answer: the mass flow rate of concentrated brine out of the process is 46,666.669 kg/hr

Explanation:

F, W and B are the fresh feed, brine and total water obtained

w = 2 x 10^4 L/h

we know that

F = W + B

we substitute

F = 2 x 10^4 + B

F = 20000 + B .................EQUA 1

solute

0.035F = 0.05B

B = 0.035F/0.05

B = 0.7F

now we substitute value of B in equation 1

F = 20000 + 0.7F

0.3F = 20000

F = 20000/0.3

F = 66666.67 kg/hr

B = 0.7F

B = 0.7 * F

B = 0.7 * 66666.67

B = 46,666.669 kg/hr

the mass flow rate of concentrated brine out of the process is 46,666.669 kg/hr

8 0
3 years ago
The capacity of a battery is 1800 mAh and its OCV is 3.9 V. a) Two batteries are placed in series. What is the combined battery
Lynna [10]

Answer:

capacity  = 0.555 mAh

capacity  = 3600 mAh

Explanation:

given data

battery = 1800 mAh

OCV = 3.9 V

solution

we get here capacity when it is in series

so here Q = 2C  

capacity  = 2 × ampere × second   ...............1

put here value and we get

and 1 Ah = 3600 C

capacity  = \frac{2}{3600}

capacity  = 0.555 mAh

and

when it is in parallel than capacity will be

capacity = Q1 +Q2   ...............2

capacity  = 1800 + 1800

capacity  = 3600 mAh

3 0
3 years ago
Estimate the endurance strength, Se, of a 37.5-mm- diameter rod of AISI 1040 steel having a machined finish and heat-treated to
7nadin3 [17]

Answer:

endurance length is 236.64 MPa

Explanation:

data given:

d = 37.5 mm

Sut = 760MPa

endurance limit is

Se = 0.5 Sut

   = 0.5*760 = 380 MPa

surface factor is

Ka = a*Sut^b

where

Sut is ultimate strength

for AISI 1040 STEEL

a = 4.51, b = -0.265

Ka = 4.51*380^{-0.265}

Ka = 0.93

size factor is given as

Kb =1.29 d^{-0.17}

Kb = 0.669

Se = Sut *Ka*Kb

    = 380*0.669*0.93

Se = 236.64 MPa

therefore endurance length is 236.64 MPa

4 0
3 years ago
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