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yKpoI14uk [10]
2 years ago
14

1. Two technicians are discussing tire rotation. Technician A says that you always follow the tire-rotation procedure outlined i

n the owner's manual or online service information. Technician B says that the modified "X" rotation pattern is seldom used. Which technician is correct? A. Both Technicians A and B B. Technician B only C. Neither Technician A nor B D. Technician A only ​
Engineering
1 answer:
siniylev [52]2 years ago
8 0

Answer:

don't know

Explanation:

huhuh

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If particleboard is used as wall sheathing, the grade mark with type _____ or _____ should be stamped on it.
kramer

If particleboard is used as wall sheathing, the grade mark with type M1 or M2 should be stamped on it.

<h3>What is particle board?</h3>

Particle board is notably used as floors underlayment or as a base for parquet floors, timber floors, or for carpets. For this purpose, the particle forums are dealt with with unique chemical compounds and resins to cause them to water-resistant or termite proof.

Waferboard, OSB, and composite plywood, while carried out as wall sheathing, offer a nail base for software of shingle siding.

Read more about the sheathing:

brainly.com/question/5029827

#SPJ1

8 0
2 years ago
A waste stabilization pond is used to treat a dilute municipal wastewater before the liquid is discharged into a river. The infl
german

Answer:

BOD concentration at the outflow = 17.83 mg/L

Explanation:

given data

flow rate of Q = 4,000 m³/day

BOD1 concentration of Cin = 25 mg/L

volume of the pond = 20,000 m³

first-order rate constant equal = 0.25/day

to find out

What is the BOD concentration at the outflow of the pond

solution

first we find the detention time that is

detention time t = \frac{volume}{flow rate}

detention time t = \frac{20000}{4000}

detention time = 5 days

so

BOD concentration at the outflow of pond is express as

BOD concentration at the outflow = Cin ( 1 - e^{-kt} )

here k is first-order rate constant and t is detention time and Cin is BOD1 concentration

so

BOD concentration at the outflow = 25 ( 1 - e^{-0.25(5)} )

BOD concentration at the outflow = 17.83 mg/L

8 0
3 years ago
Substances A and B have retention times of 16.63 and 17.63 min, respectively, on a 30 cm column. An unretained species passes th
Svet_ta [14]

Answer:

The time required to elute the two species is 53.3727 min

Explanation:

Given data:

tA = retention time of A=16.63 min

tB=retention time of B=17.63 min

WA=peak of A=1.11 min

WB=peak of B=1.21 min

The mathematical expression for the resolution is:

Re_{s} =\frac{2(t_{B}-t_{A})}{W_{A}+W_{B} } =\frac{2*(17.63-16.63)}{1.11+1.21} =0.8621

The mathematical expression for the time to elute the two species is:

\frac{t_{2}}{t_{1}} =(\frac{Re_{B} }{Re_{s} } )^{2}

Here

ReB = 1.5

t_{2} =t_{1} *(\frac{Re_{B} }{Re_{s} } )^{2} =17.63*(\frac{1.5}{0.8621} )^{2} =53.3727min

6 0
2 years ago
Time complexity of quick short
Irina-Kira [14]

Answer:

The time complexity will be "O(n log n)".

Explanation:

  • Many realistic Quick sort implementations choose a randomized special edition. The time complexity variable O(n Logn) was predicted in the randomized edition.
  • Throughout the randomized version, probably the most disgusting case is also conceivable, but by far the worst scenario for something like a given pattern does not exist as well as randomized Quick sort performs well throughout the practice.
3 0
2 years ago
A 5cm diameter copper sphere (of density = 8954 kg/m3, specific heat capacity = 0.3831 kJ/kg K) is initially at a uniform temper
Korolek [52]

Answer:

Temperature inside sphere after 10 minutes = 19924.33K

Explanation:

Detailed explanation and calculation is shown in the image below

6 0
3 years ago
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