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White raven [17]
3 years ago
10

A torque T 5 3 kN ? m is applied to the solid bronze cylinder shown. Determine (a) the maximum shearing stress, (b) the shethe 1

5-mm radiusaring stress at point D, which lies on a 15-mm-radius circle drawn on the end of the cylinder, (c) the percent of the torque carried by the portion of the cylinder within .​
Engineering
1 answer:
Lelechka [254]3 years ago
4 0

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The best way to check the efficiency of individual cylinders is:________.
KATRIN_1 [288]

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To run. The machine one at a time

Explanation:

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4 years ago
Which of the following qualities of large man-made structures is unique to dams? They span water bodies They generate electricit
Anna11 [10]

They generate electricity.

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3 years ago
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Two materials are considered _______ if one material’s property degraded another.
Natalka [10]

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Fermentation?

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3 years ago
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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  

5 0
3 years ago
If the atomic radius of copper is 0.128 nm, calculate the volume of its unit cell in cubic meters.
Alex_Xolod [135]

Answer:

Volume of face centered cubic cell=4.74531*10^{-29} m^3

Explanation:

Consider the face centered cubic cell:

1 atom at each corner of cube.

1 atom at center of each face.

Consider the one face (ABCD) as shown in attachment for calculation:

Length of the all sides of face centered cubic cell is L.

Volume of face centered cubic cell= L^3

Now Consider the figure shown in attachment:

According to Pythagoras theorem on ΔADC.

L^{2}+L^2=(4a)^2     (a is the atomic radius)

L=\frac{4a}{\sqrt{2}} (Put in the formula of Volume)

Volume of face centered cubic cell= L^3

Volume of face centered cubic cell= (\frac{4a}{\sqrt{2}})^3

Volume of face centered cubic cell= (\frac{4(0.128*10^{-9}}{\sqrt{2}})^3

Volume of face centered cubic cell=4.74531*10^{-29} m^3

3 0
4 years ago
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