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dsp73
2 years ago
12

A 1.0 kg object moving at 4.5 m/s has a wavelength of:

Physics
1 answer:
Lisa [10]2 years ago
7 0
By wave particle  duality.

Wavelength , λ = h / mv

where h = Planck's constant = 6.63 * 10⁻³⁴ Js,  m = mass in kg,  v = velocity in m/s.
m = 1kg,  v = 4.5 m/s

λ = h / mv

λ = (6.63 * 10⁻³⁴) /(1*4.5)

λ ≈  1.473 * 10⁻³⁴  m

Option D.
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A force Fof 40000 lbf is applied to rod AC the negative Y-direction. The rod is 1000 inches tall. A Force P of 25 lbf is applied
erastova [34]

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The answer is "effective stress at point B is 7382 ksi "

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Calculating the value of Compressive Axial Stress:

\to \sigma y  =\frac{F}{A} = \frac{4 F}{( p d ^2 )} = \frac{(4 x ( - 40000 \ lbf))}{[ p \times (1 \ in)^2 ]} = - 50.9 \ ksi \\

Calculating Shear Transverse:

\to \frac{4v}{ 3 A} = \frac{4 (75 \ lbf + 25 \ lbf)}{ \frac{3 ( lni)^2}{4}}

        = \frac{4 (100 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\ = \frac{400 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\= 0.17 \ ksi

= R \times 200 \ in - P \times 100 \ in = 12500 \ lbf \times\  in

\to \sigma' =[ s y^2 +3( t \times y^2 + t yz^2 )] \times \frac{1}{2}\\\\

       = [ (-50.9)^2 +3((63.7)^2 +(0.17)^2 )] \times \frac{1}{2}\\\\=[2590.81+ 3(4057.69)+0.0289]\times \frac{1}{2}\\\\=[2590.81+12,173.07+0.0289] \times \frac{1}{2}\\\\=14763.9089\times \frac{1}{2}\\\\ = 7381.95445 \ ksi\\\\ = 7382 \ ksi

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2 years ago
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