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spin [16.1K]
3 years ago
11

A 1000-n weight is hanging from a 2.0 m long aluminum rod. A 500-n weight is hanging from a 1.0 m long aluminum rod. The two alu

minum rods have different cross sectional areas. Which rod is under the greater tensile stress?
Physics
1 answer:
Hunter-Best [27]3 years ago
5 0

Answer:

Both rod have the same tensile stress

Explanation:

Given information,

The weight first rod, W_{1} = 1000 N

The length of first rod, l_{1} = 2.0 m

The weight second rod, W_{2} = 500 N

The length of second rod, l_{2 = 1.0 m

The equation of tensile stress, σ = \frac{F}{A}

where

σ = tensile stress (N/m^{2} or Pa)

F = Force (N)

A = Area (N/m^{2} or Pa)

so

σ1 =  \frac{W_{1} }{A_{1} }, A = 2πl

    = \frac{1000}{2\pi(2) }

    = \frac{250}{\pi } N/m^{2}

now calculate σ2

σ2 =  \frac{W_{2} }{A_{2} }

     = \frac{500}{2\pi(1) }

     = \frac{250}{\pi } N/m^{2}

σ1/σ2 = \frac{250}{\pi } / \frac{250}{\pi }

σ1/σ2 = 1

σ1 = σ2

Hence, the tensile stress of first and second rod are the same.

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3 years ago
A laser emits two wavelengths (λ1 = 420 nm; λ2 = 630 nm). When these two wavelengths strike a grating with 450 lines/mm, they pr
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A) Order of the first laser: 3, order of the second laser: 2

B) The overlap occurs at an angle of 34.9^{\circ}

Explanation:

A)

The formula that gives the position of the maxima (bright fringes) for a diffraction grating is

d sin \theta = m \lambda

where

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\lambda is the wavelength of the light

For laser 1,

d sin \theta = m_1 \lambda_1

For laser 2,

d sin \theta = m_2 \lambda_2

where

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Since the position of the maxima in the two cases overlaps, then the term d sin \theta on the left is the same for the two cases, therefore we can write:

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Therefore:

m_1 = 3

m_2 = 2

B)

In order to find the angle at which the overlap occurs, we use the 1st laser situation:

d sin \theta = m_1 \lambda_1

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N = 450 lines/mm = 450,000 lines/m is the number of lines per unit length, so the spacing between the lines is

d=\frac{1}{N}=\frac{1}{450,000}=2.2\cdot 10^{-6} m

m_1 = 3 is the order of the maximum

\lambda_1 = 420 nm = 420\cdot 10^{-9} m is the wavelength of the laser light

Solving for \theta, we find the angle of the maximum:

sin \theta = \frac{m_1 \lambda_1}{d}=\frac{(3)(420\cdot 10^{-9})}{2.2\cdot 10^{-6}}=0.572

So the angle is

\theta=sin^{-1}(0.572)=34.9^{\circ}

Learn more about diffraction:

brainly.com/question/3183125

#LearnwithBrainly

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