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Ahat [919]
3 years ago
11

A stretch of 1 km new concrete pavement is being laid in a 100 mm thick layer. The ambient temperature when laying the concrete

is 20°C, and you can assume that the concrete does not have any internal stress at this temperature. The thermal expansion coefficient of concrete is 10μm/m°C, and its elastic modulus is 20 GPa. Suppose all deformation of the concrete is restrained (the concrete cannot move), and suppose that the concrete behaves as a linear elastic material. Knowing its tensile strength is 10 MPa, at what temperature does this concrete crack?
Physics
1 answer:
xxMikexx [17]3 years ago
7 0

Answer:

70 °C

Explanation:

Note : 1 GPA = 1000MPa ; 1 μ = 10⁻⁶

Strain (∈) = δl / l = ∝ ΔT

Elastic modulus = Direct stress (strength)/Strain

         20 * 10³  = 10 / ∝ ΔT

ΔT  =  10 /  20 * 10³ ∝

     = 10 / ( 20 * 10³ *10⁻⁶)

      =  10/0.02 = 50

      = 50 °C

Therefore Temperature for concrete crack = ambient temperature + ΔT

20°C + 50 °C

                                                                        = 70 °C

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Explanation:

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3 0
4 years ago
On earth, you swing a simple pendulum in simple harmonic motion with a period of 1.6 seconds. what is the period of this same pe
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The period of a simple pendulum is given by
T= 2 \pi  \sqrt{ \frac{L}{g} }
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g is the acceleration of gravity

If we move the same pendulum from Earth to the Moon, its length L remains the same, while the acceleration of gravity g changes. So we can write the period of the pendulum on Earth as:
T_e= 2 \pi \sqrt{ \frac{L}{g_e} }
where g_e is the acceleration of gravity on Earth, while the period of the pendulum on the Moon is
T_m= 2 \pi \sqrt{ \frac{L}{g_m} }
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If we do the ratio of the two periods, we get
\frac{T_m}{T_e} =  \sqrt{ \frac{g_e}{g_m} }
but the gravity acceleration on the Moon is 1/6 of the gravity acceleration on Earth, so we can write g_e = 6 g_m and we can rewrite the previous ratio as
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8 0
4 years ago
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Answer:

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Explanation:

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The frictional force acting on the box, f = 30 N

The normal force on the box, η = mg

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                                                     = 490 N

The coefficient of friction,

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Hence, the coefficient of static friction between the box and floor is, μ = 0.061

7 0
4 years ago
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Answer:

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7 0
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Answer:

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