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ahrayia [7]
3 years ago
6

A photon with wavelength λ = 0.0830 nm is incident on an electron that is initially at rest. if the photon scatters in the backw

ard direction, what is the magnitude of the linear momentum of the electron just after the collision with the photon?
Physics
1 answer:
QveST [7]3 years ago
7 0
Compton scattering equation of wavelengths 
λ'-λ = h/mec (1 - CosФ)
λ' = λ + h/mec (1 - cos 180°)
= ( 0.0830nm) + (6.626 × 10⁻³⁴ J.s)/ (9.1 × 10⁻³¹ kg)(3.0 × 10⁸ m/s
= 0.0830nm
The momentum of electron is
P photon λ = Pe + P phpton λ'
Pe = h/λ - ( -h/λ') = h(λ' + λ)/λλ'
= (6.626 × 10⁻³⁴ J.s)(( 0.08785nm) +( 0.0883nm)/0.08785nm)( 0.083nm)
1.55 × 10⁻²³ kg.m/s.
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Explanation:

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V under water = A*h

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Using Newton 2nd Law

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What form of energy is a bonfire and a bunsen burner?
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3 years ago
On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s
dsp73

Answer:

(a) 1.21 m/s

(b) 2303.33 J, 152.27 J

Explanation:

m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s

(a) Let their velocity after striking is v.

By use of conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v

v = ( - 356.25 + 607.94) / 208 = 1.21 m /s

(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2

                                               = 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)

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Kinetic energy after collision = 1/2 (m1 + m2) v^2                

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Answer: 32.65\ m

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As the car is moving, so the coefficient of kinetic friction comes into play

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Using the equation of motion v^2-u^2=2as\\

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Answer:

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Explanation:

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