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Verizon [17]
2 years ago
5

a vertical solid steel post 29cm in diameter and 2.0m long is required to support a load of 8200kg, ignore the weight of the pos

t. determine the stress in the post​
Physics
1 answer:
erik [133]2 years ago
7 0

Answer:

The stress is  \sigma  =  1.218*10^{6} \  N/m^2

Explanation:

From the question we are told that

   The diameter of the post is  d =  29 \ cm  =  0.29 \  m

   The length is L  =  2.0 \  m

    The weight of the loading mass

Generally  the  radius of the post is mathematically represented as

     r =  \frac{0.29}{2}

=>   r = 0.145 \  m

Generally the area of the post is  

       A =  \pi r^2

=>     A =  3.14 *  0.145 ^2

=>     A =  0.066 \ m^2

Generally the weight exerted by the load is mathematically represented as

        F =  m  *  g

=>      F =  8200  *  9.8

=>      F =  80360 \  N

Generally the stress is mathematically represented as

         \sigma  =  \frac{F}{A}

=>      \sigma  =  \frac{80360 }{0.066}

=>      \sigma  =  1.218*10^{6} \  N/m^2

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Answer:

3.64×10⁸ m

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Let's define some variables:

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When the gravitational fields become equal:

GM₁m / r₁² = GM₂m / r₂²

M₁ / r₁² = M₂ / r₂²

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M₁ / r² = M₂ / (d² − 2dr + r²)

M₁ (d² − 2dr + r²) = M₂ r²

M₁d² − 2dM₁ r + M₁ r² = M₂ r²

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Solving with quadratic formula:

r = [ 2d ± √(4d² − 4 (1 − M₂/M₁) d²) ] / 2 (1 − M₂/M₁)

r = [ 2d ± 2d√(1 − (1 − M₂/M₁)) ] / 2 (1 − M₂/M₁)

r = [ 2d ± 2d√(1 − 1 + M₂/M₁) ] / 2 (1 − M₂/M₁)

r = [ 2d ± 2d√(M₂/M₁) ] / 2 (1 − M₂/M₁)

When we plug in the values, we get:

r = 3.64×10⁸ m

If the moon wasn't there, the acceleration due to Earth's gravity would be:

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The energy of microwaves is less than the energy of ultraviolet light. Which comparison of the energies of
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Answer:

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Explanation:

Electromagnetic waves are waves produced by the interaction between both magnetic and electric fields. These waves have some properties that make them to be arranged in a definite form producing an electromagnetic spectrum.

The spectrum has a general property of which as the wavelength of the waves increases, the frequency decreases. And vice versa.

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