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3 years ago

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A proton, an electron, and a helium nucleus all move at speed v. Rank their de Broglie wavelengths from largest to smallest. Wri

te any relevant formulas you use below. a) helium nucleus, proton, electron
b) helium nucleus, electron, proton
c) electron, proton, helium nucleus
d) proton, electron, helium nucleus
e) proton, helium nucleus, electron

1 answer:

MatroZZZ [7]3 years ago

6 0

Answer:

The correct option is 'c':electron,proton,helium nucleus

Explanation:

The De-Broglie's wavelength of particle is given by

$\lambda =\frac{h}{mv}$

Thus we can see that wavelength is inversely related to mass of the particle since 'h' (Plank's constant) and velocity is same for all the particles

Thus we conclude that the the lightest particle will have the most wavelength

Electron being the lightest of the 3 particles will have the largest wavelength thus the correct option is 'c'. Since electron has the largest wavelength followed by proton and the least wavelength among the 3 is of helium.

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A crate resting on a rough horizontal floor is to be moved horizontally. The coefficient of static friction is 0.36. To start th

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Answer:

Explanation:

The direction of force will be in upward direction making an angle of θ with the vertical .

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= F sinθ

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F sinθ = .36 (mg - F cosθ)

F sinθ + .36 F cosθ =.36 mg

F (sinθ + .36 cosθ) = .36 mg

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F R sin( θ+δ ) = . 36 mg

F = .36 mg / Rsin( θ+δ )

For minimum F , sin( θ+δ ) should be maximum

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3 years ago

When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0

lora16 [44]

Answer:

$d=0.165m$

Explanation:

Given

$m=0.25kg$ , $x_{1}=5cm*\frac{1m}{100cm}=0.05m$ , $x_{2}=4cm*\frac{1m}{100cm}=0.04m$ , $v=2\frac{rev}{s}$

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$K=\frac{m*g}{x_{1}}$

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$F_{c}=\frac{m*v^2}{r}$

$v=w*r=2\pi*r$

But Fc is also

Fc=KxΔr

$F_{c}=K*(r-x_{2})$

Replacing

$m*4\pi^2*r=K*(r-x_{2})$

$0.25kg*4\pi^2*r=49*(r-0.04m)$

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3 years ago

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Explanation:

Given

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