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3 years ago

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A hanging wire made of an alloy of titanium with diameter 0.05 cm is initially 2.7 m long. When a 15 kg mass is hung from it, th

e wire stretches an amount 1.68 cm. A mole of titanium has a mass of 48 grams, and its density is 4.54 g/cm3. Based on these experimental measurements, what is Young's modulus for this alloy of titanium

1 answer:

Svet_ta [14]3 years ago

8 0

Answer:

Explanation:

Young's modulus of elasticity Y = stress / strain

stress = force / cross sectional area

= weight of 15 kg / π r²

= 15 x 9.8 / 3.14 x ( .025 x 10⁻² )²

stress = 74.9 x 10⁷ N / m²

strain = Δ L / L , Δ L is change in length and L is original length

Putting the values

strain = .0168 / 2.7 =.006222

Young's modulus of elasticity Y = 74.9 x 10⁷ / .006222

= 120.88 x 10⁹ N / m² .

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Answer:

Part A:

The proton has a smaller wavelength than the electron.

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Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

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The wavelength of each particle can be determined by means of the De Broglie equation.

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$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

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<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

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Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

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3 years ago

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