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3 years ago

15

Determine the percent increase in the nominal moment capacity of the section in Problem 2 when including compression steel at to

p equal to 0.5 the area of the tension steel at the bottom.

1 answer:

sergejj [24]3 years ago

5 0

Explanation:

Please kindly share your problem two with us as to know the actual problem we are dealing with, the question looks incomplete

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If the contact surface between the 20-kg block and the ground is smooth, determine the power of force F when t = 4 s. Initially,

grigory [225]

Answer:

The power of force F is 115.2 W

Explanation:

Use following formula

Power = F x V

$F_{H}$ = F cos0

$F_{H}$ = (30) x 4/5

$F_{H}$ = 24N

Now Calculate V using following formula

V = $V_{0}$ + at

$V_{0}$ = 0

a = $F_{H}$ / m

a = 24N / 20 kg

a = 1.2m / $S^{2}$

no place value in the formula of V

V = 0 + (1.2)(4)

V = 4.8 m/s

So,

Power = $F_{H}$ x V

Power = 24 x 4.8

Power = 115.2 W

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3 years ago

The critical resolved shear stress for iron is 27 MPa (4000 psi). Determine the maximum possible yield strength for a single cry

jasenka [17]

Answer:

Answer is mentioned below.

Explanation:

Answer: Calculation of maximum yield strength for a single crystal of Fe pulled in tension is shown in the image attached .

value of max. Yield strength= 54MPa

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4 years ago

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mart [117]

Answer:

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3 years ago

Which of the following are some of the problems found in city streets?

storchak [24]

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3 years ago

Base course aggregate has a target dry density of 119.7 lb/cu ft in place. It will be laid down and compacted in a rectangular s

natita [175]

Answer:

total weight of aggregate = 5627528 lbs = 2814 tons

Explanation:

given data

dry density = 119.7 lb/cu ft

area = 2000 ft × 48 ft × 6 in

aggregate = 3.1%

required compaction = 95%

solution

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

volume = 2000 × 48 × 0.5 = 48000 ft³

when here space fill with aggregate of density is

density = 0.95 × 119.7 = 113.72 lb/ft³

and

dry weight of this aggregate will be is

dry weight = 48000 × 113.72 = 5458320 lbs

and

we consider take percent moisture by weigh so that there weight of moisture in aggregate is express as

weight of moisture = 0.031 × 5458320 = 169208 lbs

and

total weight of aggregate will be

total weight of aggregate = 5458320 + 169208

total weight of aggregate = 5627528 lbs = 2814 tons

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3 years ago