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Elan Coil [88]
3 years ago
8

A steel rod is pulled in tension with a stress that is less than the yield strength. The modulus of elasticity may be calculated

as : axial stress divided by change in length.
axial stress divided by axial strain.
axial stress times axial strain.
axial load divided by change in length.
Physics
1 answer:
pshichka [43]3 years ago
6 0

Answer:

B. Axial stress divided by axial strain

Explanation:

Elasticity:

It is the tendency of an object to deform along the axis when an opposing force is applied without facing permanent change in shape.

Plasticity:

When an object crosses the elasticity limit, it enters plasticity where the change due to stress is permanent and the object might even break.

Yield strength:

Yield strength is the point of maximum bearable stress that indicates the limit of elasticity.

Our case:

As the stress applied is less than the yield strength, the rod is still in the elasticity state and its modulus can be calculated.

Modulus of Elasticity = Stress along axis/Ratio of change in length to original length

Axial strain is basically the ratio of change in length to original length.

So, Modulus of Elasticity = Axial Stress/ Axial Strain

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A 1300 kg steel beam is supported by two ropes. (Figure
Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

  • the net horizontal force acting on the beam is

R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0

where R_1,R_2 are the magnitudes of the tensions in ropes 1 and 2, respectively;

  • the net vertical force acting on the beam is

R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0

where m=1300\,\rm kg and g=9.8\frac{\rm m}{\mathrm s^2}.

Eliminating R_2, we have

\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)

R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2

R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2

-R_1 \sin(50^\circ) = -\dfrac{mg}2

R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}

Solve for R_2.

\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0

\dfrac{R_2}2 = -mg\cot(110^\circ)

R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}

8 0
1 year ago
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A car sitting at a red light begins to accelerate at 2 m/s2 when the light turns green. It continues with this acceleration unti
jek_recluse [69]

Answer:

    x_total = 600 m

Explanation:

This is an exercise and kinematics, let's find the time it takes to reach the velocity  20 m / s

          v = v₀ + a t

           

as part of rest v₀ = 0

          t = v / a

          t = 20/2

          t = 10 s

let's find the distance traveled in this time

          x₁ = vo t + ½ a t2

          x₁ = 0 + ½ 2 10²

          x₁ = 100 m

The remaining time is

          t₂ = 35 - t

          t₂ = 35 - 10

          t₂ = 25 s

as in this range it has a constant speed

          v = x₂ / t₂

          x₂ = v t₂

          x₂ = 20  25

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the total distance traveled is

           x_total = x₁ + x₂

           x_total = 100 + 500

           x_total = 600 m

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Answer:

6.21 m/s

Explanation:

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U_{1-2}=T_B- T_A\\58d-mgh=0.5m(v_b^{2}-v_a^{2})

where d is displacement from initial to final position, v is velocity and subscripts a and b are position A and B respectively, m is mass of collar, g is acceleration due to gravity

Substituting 1 Kg for m, 0.4m for h, v_a as 0, 9.81 for g then

58(\sqrt{0.4^{2}+0.3^{2}}-0.1)-(1\times 9.81\times 0.4)=0.5\times 1\times (v_b^{2}-v_a^{2})\\19.276=0.5\times 1v_b^{2}\\v_b=6.209025688 m/s\approx 6.21 m/s

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