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Tema [17]
3 years ago
12

An acetal polymer block is fixed to the rigid plates at its top and bottom surfaces. If the top plate displaces 2 mm horizontall

y when it is subjected to a horizontal force P = 2 kN ,determine the shear modulus of the polymer. The width of the block is 100 mm. Assume that the polymer is linearly elastic and use small angle analysi
Physics
1 answer:
Taya2010 [7]3 years ago
3 0

Answer:

The shear modulus of the polymer is 5 MPa

Solution:

As per the question:

displacement of the top plate, d = 2 mm = 0.002 m

Horizontal Force, P = 2 kN = 2000 N

Width of the block, w = 100 mm = 0.1 m

By small angle analysis:

Angle, \gamma = \frac{2}{200} = 0.01 rad

Now,

Stress, \tau = \frac{P}{Area} = \frac{2000}{0.4\times 0.1} = 50000 Pa

G = \frac{\tau}{\gamma} = \frac{50000}{0.01} = 5 MPa

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Answer:

They had different beliefs and backgrounds.  

Explanation:

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Two electrons in a vacuum exert force of F = 3.8E-09 N on each other. They are then moved such that they are separated by x = 8.
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Answer:

F_n = 5.65E-11 N

d =  1.20682E-31 m

Explanation:

F = 3.8E-09 N

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x = Distance between them

F=G\frac{m^2}{x^2}\\\Rightarrow 3.8E-09=G\frac{m^2}{x^2}

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F_n=G\frac{m^2}{x^2}\\\Rightarrow F_n=G\frac{m^2}{(8.2x)^2}\\\Rightarrow F_n=G\frac{m^2}{67.24x^2}

Dividing the above equations we get

\frac{F}{F_n}=\frac{G\frac{m^2}{x^2}}{G\frac{m^2}{67.24x^2}}\\\Rightarrow \frac{F}{F_n}=67.24\\\Rightarrow F_n=\frac{F}{67.24}\\\Rightarrow F_n=\frac{3.8E-09}{67.24}\\\Rightarrow F_n=5.65E-11\ N

F_n = 5.65E-11 N

F=G\frac{m^2}{x^2}\\\Rightarrow x=\sqrt{\frac{Gm^2}{F}}\\\Rightarrow x=\sqrt{\frac{G}{F}}m\\\Rightarrow x=\sqrt{\frac{6.67E-11}{3.8E-09}}9.109E-31\\\Rightarrow x=1.20682E-31\ m

d =  1.20682E-31 m

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