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3 years ago

6

A circular rod with a gage length of 3.5 mm and a diameter of 2.8 cmcm is subjected to an axial load of 68 kN . If the modulus o

f elasticity is 200 GPa, what is the change in length?

1 answer:

Crank3 years ago

7 0

To solve this problem, we will apply the concepts related to the linear deformation of a body given by the relationship between the load applied over a given length, acting by the corresponding area unit and the modulus of elasticity. The mathematical representation of this is given as:

$\delta = \frac{PL}{AE}$

Where,

P = Axial Load

l = Gage length

A = Cross-sectional Area

E = Modulus of Elasticity

Our values are given as,

l = 3.5m

D = 0.028m

$P = 68*10^3 N$

E = 200GPa

$A = \frac{\pi}{4}(0.028)^2 \rightarrow 0.0006157m^2$

Replacing we have,

$\delta = \frac{PL}{AE}$

$\delta = \frac{( 68*10^3)(3.5)}{(0.0006157)(200*10^9)}$

$\delta = 0.001932m$

$\delta = 1.93mm$

Therefore the change in length is 1.93mm

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2 years ago

Green light (λ = 518 nm) strikes a single slit at normal incidence. What width slit will produce a central maximum that is 3.00

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Answer:

6.9066 × 10⁻⁵ m

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The formula can be written as:

$sin\theta=\frac {\lambda}{d}\times m$ ....1

The location of the bright fringe is determined by :

$y=L\times tan\theta$

Where, L is the distance between the slit and the screen.

For small angle , $sin\theta=tan\theta$

So,

Formula becomes:

$y=L\times sin\theta$

Using 1, we get:

$y=L\times \frac {\lambda}{d}\times m$

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1 cm = 0.01 m

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λ = 518 × 10⁻⁹ m

Applying the formula as:

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3 years ago

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Answer:

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