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stellarik [79]
3 years ago
11

). A company periodically tests its product for tread wear under simulated conditions. Thirty random samples, each containing 5

items, were taken in a control chart application and it was found that the grand average is

Engineering
1 answer:
Luden [163]3 years ago
5 0

Answer:

See explanation for step by step answer.

Explanation:

Given that:

A company periodically tests its product for tread wear under simulated conditions. Thirty random samples, each containing 5 items, were taken in a control chart application and it was found that the grand average is....

See attacent for complete answer

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List and explain 4 factors you need to observe while stick welding to make a good “consistent” bead
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Answer:I don’t know this one

Explanation:

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3 years ago
Base course aggregate has a target dry density of 119.7 Ib/cu ft in place. It will be laid down and compacted in a rectangular s
djyliett [7]

Answer:

total weight of aggregate =  5627528 lbs = 2814 tons  

Explanation:

we get  here volume of space to be filled with aggregate that is

volume = 2000 × 48 × 0.5

volume = 48000 ft³

now filling space with aggregate of the density that is

density = 0.95 × 119.7

density = 113.72 lb/ft³

and dry weight of this aggregate is

dry weight = 48000 × 113.72

dry weight = 5458320 lbs

we consider here percent moisture is by weigh

so weight of moisture in aggregate will be

weight of moisture = 0.031 × 5458320

weight of moisture = 169208 lbs

so here total weight of aggregate is

total weight of aggregate = 5458320 + 169208

total weight of aggregate =  5627528 lbs = 2814 tons  

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2 years ago
For an isotropic material, E and ν are often chosen as the two independent engineering constants. There are other elastic consta
pav-90 [236]

Answer:

khgy

Explanation:

nbv

8 0
3 years ago
THIS SIGN MEANS: A. The right lane will end ahead B. The highway will be divided ahead C. Less space between lanes ahead
hodyreva [135]
B.) the highway will be divided ahead
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An unknown immiscible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the unknown
barxatty [35]

Answer:

The specific weight of unknown liquid is found to be 15 KN/m³

Explanation:

The total pressure in tank is measured to be 65 KPa in the tank. But, the total pressure will be equal to the sum of pressures due to both oil and unknown liquid.

Total Pressure = Pressure of oil + Pressure of unknown liquid

65 KPa = (Specific Weight of oil)(depth of oil) + (Specific Weight of unknown liquid)(depth of unknown liquid)

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(Specific Weight of Unknown Liquid)(1.5 m) = 65 KN/m² - 42.5 KN/m²

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<u>Specific Weight of Unknown Liquid = 15 KN/m³</u>  

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3 years ago
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