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denis23 [38]
3 years ago
11

Amazon rectangular bar of low carbon steel A-36 is exposed to an axial strees of 150 MPa. What is the original length of the bar

in m, if it elongates 1.35 mm under these conditions?
Physics
1 answer:
kolbaska11 [484]3 years ago
5 0

Answer:

1.8m

Explanation:

Let the Elastics of the steel ASTM-36 E = 200000 MPa

The strain of the bar when subjected to 150 MPa is

\epsilon = \frac{\sigma}{E} = \frac{150}{200000} = 0.00075

Therefore, if the bar elongates by 1.35 mm, then the original length L would be:

\epsilon = \frac{\Delta L}{L}

L = \frac{\Delta L}{\epsilon} = \frac{1.35}{0.00075} = 1800 mm or 1.8m

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Write an essay about Paralympic and Special Olympic Essay 100 points
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The Paralympics does the same. The Special Olympics and the Paralympics are both sports organizations for disabled athletes, but they are very different organizations and are not connected. Many people confuse the Paralympics and the Special Olympics.

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A man is dragging a trunk up the loading ramp of a mover's
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Horizontal component = 241 N

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Explanation:

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Force =  F = 375 N

Referring to diagram attached, the force F is making an angle

theta = 20+30 = 50 with the horizontal.

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An object with a mass of 0. 25 kg is undergoing simple harmonic motion at the end of a vertical spring with a spring constant, k
skelet666 [1.2K]

Answer:

1) The amplitude of the motion is approximately 0.274 meters.

2) The total energy of the object at any point of its motion is 16.892 joules.

Explanation:

1) An object under simple harmonic motion is conservative, since there is no dissipative forces acting during motion (i.e. friction, air viscosity). The amplitude of the motion can be found easily by Principle of Energy Conservation by the fact that maximum elastic potential energy (U_{e}), in joules, is equal to maximum translational kinetic energy (K), in joules:

U_{e} = K

\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2} (1)

Where:

k - Spring constant, in newtons per meter.

A - Amplitude, in meters.

m - Object mass, in kilograms.

v - Speed of the object at equilibrium, in meters per second.

If we know that k = 450\,\frac{N}{m}, m = 0.25\,kg and v = 0.3\,\frac{m}{s}, then the amplitude of the motion is:

\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2}

k\cdot A^{2} = m\cdot v^{2}

A = v\cdot \sqrt{\frac{m}{k} }

A = \left(0.3\,\frac{m}{s} \right)\cdot \sqrt{\frac{0.25\,kg}{0.3\,\frac{m}{s} } }

A \approx 0.274\,m

The amplitude of the motion is approximately 0.274 meters.

2) The total energy of the object (E), in joules, is found either by maximum elastic potential energy or by maximum translational kinetic energy, that is: (k = 450\,\frac{N}{m}, A \approx 0.274\,m)

E = U_{e}

E = \frac{1}{2}\cdot k\cdot A^{2}

E = \frac{1}{2}\cdot \left(450\,\frac{N}{m} \right) \cdot (0.274\,m)^{2}

E = 16.892\,J

The total energy of the object at any point of its motion is 16.892 joules.

7 0
3 years ago
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