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Alex73 [517]
3 years ago
13

A uniform steel rod of cross-sectional area A is attached to rigid supports and is unstressed at a temperature of 458F. The stee

l is assumed to be elastoplastic with sY 5 36 ksi and E 5 29 3 106 psi. Knowing that a 5 6.5 3 1026 /8F, determine the stress in the bar (a) when the temperature is raised to 3208F, (b) after the temperature has returned to 458F.
Physics
2 answers:
Vedmedyk [2.9K]3 years ago
6 0

Explanation:

As the given rod is attached to rigid supports as a result, the deformation occurring due to the change in temperature will cause stress in the rod.

Let us assume that P is the compressive force in the rod due to change in temperature.

So,        \Delta T = \frac{\sigma_{y}}{E_{a}}

                          = \frac{36 \times 10^{3}}{(29 \times 10^{6} \times 6.5 \times 10^{-6})}

                          = 190.98^{o}F

Now, we will calculate the actual change in temperature as follows.

             \Delta T = 320 - 45 = 275^{o}F

This means that the actual change in temperature is more than required for yielding.

(a)   Formula to calculate yielding stress is as follows.

             \sigma' = \frac{P'}{A}

           \sigma' = -\frac{AE \alpha \Delta T}{A}

                        = -E \alpha \Delta T

                        = -29 \times 10^{6} \times 6.5 \times 10^{-6} \times 275

                        = -51.8375 \times 10^{3} psi

Hence, stress in the bar when temperature is raised to 320^{o}F is -51.8375 \times 10^{3} psi.

(b)  Now, we will calculate the residual stress as follows.

            \sigma_{r} = -\sigma_{y} - \sigma'

           \sigma_{r} = -36 + 51.837 ksi

                          = 15.837 ksi

Therefore, stress in the bar when the temperature has returned to 45^{o}F is 15.837 ksi.

artcher [175]3 years ago
4 0

Answer:

(a). The stress is -51.83\times10^{3}\ psi

(b). The residual stress is 15.83 ksi

Explanation:

Given that,

Initial temperature = 45°F

Raised temperature = 320°F

\sigma_{y}=36\ ksi

E=29\times10^{6}\ psi

\alpha=6.5\times10^{-6}\ /^{\circ}F

We need to calculate the actual change in temperature

\Delta T_{a}=320-45=275^{\circ}F

We need to calculate the stress

Using formula of stress

\sigma'=\dfrac{P}{A}

\sigma'=-\dfrac{AE\alpha\Delta T}{A}

\sigma'=-E\alpha\Delta T

Put the value into the formula

\sigma'=-29\times10^{6}\times6.5\times10^{-6}\times275

\sigma'=-51.83\times10^{3}\ psi

(b). We need to calculate the residual stress

Using formula of stress

\sigma_{r}=-\sigma_{y}-\sigma'

Put the value into the formula

\sigma_{r}=-36+51.83\times10^{3}

\sigma_{r}=15.83\ ksi

Hence, (a). The stress is -51.83\times10^{3}\ psi

(b). The residual stress is 15.83 ksi

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Answer:

Explanation:

The variables we know and are given are:

time, t = 20s

Charge, Q = 3x1-^-6 electrons, which is just 3x10^-6C (C stands for Coulombs, which is the unit for Charge)

We need to find the current, I, and since we know Q and t we can substitute these values into the given equation:

I=Q/t (which if you look at what the RHS is saying, its Charge over time, or more literally means the amount of charge passing a point over a period of time)

If we substitute these values, we will get I as:

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3 years ago
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The concept required to solve this problem is linked to inductance. This can be defined as the product between the permeability in free space by the number of turns squared by the area over the length. Recall that Inductance is defined as the opposition of a conductive element to changes in the current flowing through it. Mathematically it can be described as

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Replacing with our values we have,

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How can we realize that light travel in straight line ?​
Norma-Jean [14]

Answer:

It can be seen from the operation of pin-hole camera, formation of shadows and eclipse.

Explanation:

The phenomenon of light traveling in a straight line is known as rectilinear propagation of light.

One this evidence can be seen from the operation of pin-hole camera, which depends on rectilinear propagation of light

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0.00970 s

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Force due to magnetic field = qvB sin θ

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v = velocity of the charge

B = magnetic field strength = 2.7 T

θ = angle between the velocity of the charge and the magnetic field = 90°, sin 90° = 1

F = qvB

Centripetal force responsible for circular motion = mv²/r = mvw

where w = angular velocity.

The centripetal force that causes the charge to move in a circular motion = The force exerted on the charge due to magnetic field

mvw = qvB

mw = qB

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w = 3.24 × 10² rad/s

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siniylev [52]

Answer:

Explanation:

Given that,

The volume of the balloon is

V = 440 × 10³ m³

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The buoyant force is the weight of water displaced and it is calculated using

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g is the acceleration due to gravity

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