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The state of strain at a point is plane strain with εx = ε0, εy = –2ε0, γxy = 0, where ε0 is a positive constant. What is the no

Marat540 [252]

Answer:

The normal strain along an axis oriented 45° from the positive x axis in the clockwise direction is -ε₀/2

Explanation:

Given that

$\epsilon_{x}=\epsilon_{o}\\\\\epsilon_{y}=-2\epsilon_{o}\\\\\gamma_{xy}=0\\\\\theta=-45^{o}\\\\\epsilon_{x_{1}}=?$

From equation of normal strain in x direction:

$\epsilon_{x_{1}}=\epsilon_{x}cos^{2}\theta+\epsilon_{y}sin^{2}\theta+\gamma_{xy{ sin\theta cos\theta$

Substituting the values:

$\epsilon_{x_{1}}=\epsilon_{o}cos^{2}(-45)-2\epsilon_{o}sin^{2}(-45)+0\\\\\epsilon_{x_{1}}=\frac{\epsilon_{o}}{2}-2\frac{\epsilon_{o}}{2}\\\\\epsilon_{x_{1}}=-\frac{\epsilon_{o}}{2}$

6 0

3 years ago

A bicyclist of mass 60kg supplies 340W of power while riding into a 5 m/s head wind. The frontal area of the cyclist and bicycle

NikAS [45]

Answer:

87.1 mph

Explanation:

We are given that

Mass,m=60 kg

Power,P=340 W

Speed,v=5 m/s

Area, $A=0.344 m^2$

Drag coefficient, $C_d=0.88$

Coefficient of rolling resistance, $\mu_r=0.007$

Friction force, $f=\mu_rmg=0.007\times 60\times 9.8=4.1 N$

Where $g=9.8 m/s^2$

Let speed of cyclist=v'

Drag force, $F_d=\frac{1}{2}\rho_{air}AC_dv^2$

Density of air, $\rho_{air}=1.225 kg/m^3$

$F_d=\frac{1}{2}\times 1.225\times 0.344\times 0.88(5)^2=4.635N$

Power,P= $(F_d+f)\times v'$

$340=(4.1+4.635) v'=8.735v'$

$v'=\frac{340}{8.735}=38.9m/s$

$v'=87.1 mph$

1 m=0.00062137 miles

1 hour=3600 s

7 0

3 years ago

In a classroom demonstration, students are using a Slinky to observe and learn about wave properties. If the Slinky has a period

Andrej [43]

Answer:

Frequency = 3.0 Hertz

Explanation:

Given the following data;

Period = 0.333 seconds

To find the frequency;

Mathematically, frequency of a wave is given by the formula;

Frequency = 1/period

Substituting into the formula, we have;

Frequency = 1/0.333

Frequency = 3.0 Hertz

3 0

3 years ago

How can tidal force from the moon affect our earth?

VARVARA [1.3K]

Answer:

sorry but I can understand the question

5 0

3 years ago

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A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In

atroni [7]

Answer:

The average induced emf around the border of the circular region is $8.48\times 10^{-5}\ V$ .

Explanation:

Given that,

Radius of circular region, r = 1.5 mm

Initial magnetic field, B = 0

Final magnetic field, B' = 1.5 T

The magnetic field is pointing upward when viewed from above, perpendicular to the circular plane in a time of 125 ms. We need to find the average induced emf around the border of the circular region. It is given by the rate of change of magnetic flux as :

$\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=A\dfrac{-d(B'-B)}{dt}\\\\\epsilon=\pi (1.5\times 10^{-3})^2\times \dfrac{1.5}{0.125}\\\\\epsilon=8.48\times 10^{-5}\ V$

So, the average induced emf around the border of the circular region is $8.48\times 10^{-5}\ V$ .

6 0

3 years ago

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