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slega [8]
3 years ago
6

What is a shearing stress? Is there a force resulting from two solids in contact to which is it similar?

Engineering
1 answer:
Luba_88 [7]3 years ago
7 0

Answer:

Shearing stresses are the stresses generated in any material when a force acts in such a way that it tends to tear off the material.

Generally the above definition is valid at an armature level, in more technical terms shearing stresses are the component of the stresses that act parallel to any plane in a material that is under stress. Shearing stresses are present in a body even if normal forces act on it along the centroidal axis.

Mathematically in a plane AB the shearing stresses are given by

\tau =\frac{Fcos(\theta )}{A}

Yes the shearing force which generates the shearing stresses is similar to frictional force that acts between the 2 surfaces in contact with each other.  

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What is a motor cycle motor made out of
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Explanation:

a motorcycle Motor is made out of iron

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The cars of a roller-coaster ride have a speed of 19.0 km/h as they pass over the top of the circular track. Neglect any frictio
Dmitrij [34]

Complete Question

The cars of a roller-coaster ride have a speed of 19.0 km/h as they pass over the top of the circular track. Neglect any friction and calculate their speed v when they reach the horizontal bottom position. At the top position, the radius of the circular path of their mass centers is 21 m, and all six cars have the same mass.V = -18 m What is v?X km/h

Answer:

v=23.6m/s

Explanation:

Velocity v_c=18.0km/h

Radius r=21m

initial velocity uu=19=>5.27778

Generally the equation for Angle is mathematically given by

\theta=\frac{v_c}{2r}

\theta=\frac{18}{2*21}

\theta=0.45

\theta=25.7831 \textdegree

Generally

Height of mass

h=\frac{rsin\theta}{\theta}

h=\frac{21sin25.78}{0.45}

h=20.3m

Generally the equation for Work Energy is mathematically given by

0.5mv_0^2+mgh=0.5mv^2

Therefore

v=\sqrt{u^2+2gh}

v=\sqrt{=5.27778^2+2*9.81*20.3}

v=23.6m/s

3 0
3 years ago
The following are the results of a sieve analysis. U.S. sieve no. Mass of soil retained (g) 4 0 10 18.5 20 53.2 40 90.5 60 81.8
il63 [147K]

Answer:

a.)

US Sieve no.                         % finer (C₅ )

4                                                  100

10                                                95.61

20                                               82.98

40                                               61.50

60                                               42.08

100                                              20.19

200                                              6.3

Pan                                               0

b.) D10 = 0.12, D30 = 0.22, and D60 = 0.4

c.) Cu = 3.33

d.) Cc = 1

Explanation:

As given ,

US Sieve no.             Mass of soil retained (C₂ )

4                                            0

10                                          18.5

20                                         53.2

40                                         90.5

60                                         81.8

100                                        92.2

200                                       58.5

Pan                                        26.5

Now,

Total weight of the soil = w = 0 + 18.5 + 53.2 + 90.5 + 81.8 + 92.2 + 58.5 + 26.5 = 421.2 g

⇒ w = 421.2 g

As we know that ,

% Retained = C₃ = C₂×\frac{100}{w}

∴ we get

US Sieve no.               % retained (C₃ )               Cummulative % retained (C₄)

4                                            0                                           0

10                                          4.39                                      4.39

20                                         12.63                                     17.02

40                                         21.48                                     38.50

60                                         19.42                                     57.92

100                                        21.89                                     79.81

200                                       13.89                                     93.70

Pan                                        6.30                                      100

Now,

% finer = C₅ = 100 - C₄

∴ we get

US Sieve no.               Cummulative % retained (C₄)          % finer (C₅ )

4                                                     0                                          100

10                                                  4.39                                      95.61

20                                                 17.02                                     82.98

40                                                 38.50                                    61.50

60                                                 57.92                                    42.08

100                                                79.81                                     20.19

200                                                93.70                                   6.3

Pan                                                 100                                        0

The grain-size distribution is :

b.)

From the diagram , we can see that

D10 = 0.12

D30 = 0.22

D60 = 0.12

c.)

Uniformity Coefficient = Cu = \frac{D60}{D10}

⇒ Cu = \frac{0.4}{0.12} = 3.33

d.)

Coefficient of Graduation = Cc = \frac{D30^{2}}{D10 . D60}

⇒ Cc = \frac{0.22^{2}}{(0.4) . (0.12)} = 1

3 0
3 years ago
Microchips found inside most electronic devices today are examples of what material A. Polymers B. Alloys C. Composites D. None
dedylja [7]

Answer: A

Explanation:

Microchips are made out of silicone witch is a polymer.

6 0
3 years ago
Read 2 more answers
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