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Slav-nsk [51]
4 years ago
3

As stated in the introduction, shear strength is another measure of the strength of a material. A shear force is a force that ac

ts parallel to the plane in the material that breaks. A good example of a shear is that of a martial arts expert breaking boards or bricks with her hands. Other applications in which shear forces and shear strength need to be known are geology, for studying earthquakes and landslides; fluid dynamics; and structural engineering. Aluminum has a shear strength of 210 megapascals. When you bend aluminum foil around an edge (i.e., the edge of the box) and pull, you are effectively applying a shear force along the bent edge of the foil. If a roll of household aluminum foil is 30.0 centimeters wide and its thickness is approximately 15.0 micrometers, how much shear force is needed to pull off a sheet?
Physics
1 answer:
irakobra [83]4 years ago
8 0

Answer:

F=945\ N

Explanation:

Given:

  • shear stress strength of the aluminium, \tau=210\ MPa
  • width of aluminium foil, w=300\ mm
  • thickness of aluminium foil, t=15\times 10^{-3}\ mm

<u>We know the relation between shear force and shear stress as:</u>

\tau=\frac{F}{A}

where:

A = area subjected to the force F

Here the area subjected under the shear force is the bent part of the aluminium foil whole along the width.

210=\frac{F}{15\times 10^{-3} \times 300}}

F=945\ N

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3 years ago
Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consid
Korolek [52]

Answer:

a

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b

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Explanation:

From the question we are told that

     Their distance apart is  d =  5.00 \ m

      The  wavelength of each source wave \lambda =  6.0 \ m

Let the distance from source A  where the construct interference occurred be z

Generally the path difference for constructive interference is

              z - (d-z) =  m \lambda

Now given that we are considering just the straight line (i.e  points along the line connecting the two sources ) then the order of the maxima m =  0

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=>     2 z - 5 =  0

=>     z= 2.5 \ m

Generally the path difference for destructive  interference is

           |z-(d-z)| = (2m + 1)\frac{\lambda}{2}

=>         |2z - d |= (0 + 1)\frac{\lambda}{2}

=>        |2z - d| =\frac{\lambda}{2}

substituting values

          |2z - 5| =\frac{6}{2}

=>      z  =  \frac{5 \pm 3}{2}

So  

      z =  \frac{5 + 3}{2}

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and

      z =  \frac{ 5 -3 }{2}

=>   z =  1 \ m

=>    z =  (1 \ m ,  4 \ m )

7 0
3 years ago
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Now, we manipulate the given value:

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