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3 years ago

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The initial velocity of a 4.0kg box is 11m/s due west. After the box slides 4.0m horizontally its speed is 1.5 m/s. Determine th

e magnitude and the direction of the non conservative force acting on the box as it slides.

1 answer:

lorasvet [3.4K]3 years ago

6 0

Answer:

The magnitude and direction of the non conservative force acting on the box is <F= 59.36 N - DUE EAST DIRECTION>.

Explanation:

m= 4 kg

Vi= 11 m/s

Vf= 1.5 m/s

d= 4m

d= Vi * t - a * t²/2

clearing a:

a= 2*(Vi * t - d)/ t²

Vf= Vi - a * t

replacing "a" and clearing t:

t= 2d/(Vf+Vi)

t= 0.64 s

found now the value of a:

a= Vi - Vf / t

a= 14.84 m/s ²

F= m * a

F= 59.36 N

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a) de Broglie wavelength of a 4-MeV proton: 14.3 fm

b) de Broglie wavelength of a 40-GeV electron: 0.031 fm

Explanation:

a)

The de Broglie wavelength of an object is given by

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Here we want to find the de Broglie wavelength of a 4-MeV proton. The rest of mass of the proton in MeV is

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Substituting, we find

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In this case, the electron has kinetic energy of 40 GeV, while the rest mass of an electron is

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Since $40 GeV >> 0.511 MeV$ , the electron is ultra-relativistic: so we can rewrite its energy as

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3 years ago

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Answer:

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