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3 years ago

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A proton in a particle accelerator is traveling at a speed of 0.99c.(a) If you use the approximate nonrelativistic equation for

the magnitude of momentum of the proton, what answer do you get?(b) What is the magnitude of the correct relativistic momentum of the proton?(c) The approximate value (the answer to part a) is significantly too low. What is the ratio of magnitudes you calculated (correct/approximate)?

1 answer:

____ [38]3 years ago

4 0

Answer:

a) p = 4.96 10⁻¹⁹ kg m / s , b) p = 35 .18 10⁻¹⁹ kg m / s ,

c) p_correst / p_approximate = 7.09

Explanation:

a) The moment is defined in classical mechanics as

p = m v

Let's calculate its value

p = 1.67 10⁻²⁷ 0.99 3. 10⁸

p = 4.96 10⁻¹⁹ kg m / s

b) in special relativity the moment is defined as

p = m v / √(1 –v² / c²)

Let's calculate

p = 1.67 10⁻²⁷ 0.99 10⁸/ √(1- 0.99²)

p = 4.96 10⁻¹⁹ / 0.141

p = 35 .18 10⁻¹⁹ kg m / s

c) the relationship between the two values is

p_correst / p_approximate = 35.18 / 4.96

p_correst / p_approximate = 7.09

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iogann1982 [59]

Answer:

v = 0.41 m/s

Explanation:

In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
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The change in the potential energy, can be written as follows:

$\Delta U = U_{f} - U_{o} = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)$

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Regarding the work done by the force of friction, it can be written as follows:

$W_{ffr} = - \mu_{k}* F_{n} * \Delta x (4)$

where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

horizontal displacement.

Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
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Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

$\frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o}) (6)$

$\frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x) -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)$

Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:

$\frac{1}{2} * 2.00 kg* v^{2} = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)$

$v =\sqrt{(0.326-0.1568} = 0.41 m/s (9)$

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3 years ago

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2 years ago

A double-slit experiment uses light of wavelength 650 nm with a slit separation of 0.100 mm and a screen placed 4.0 m away. a) W

dezoksy [38]

Answer:

Explanation:

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3 years ago

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Answer:

(a) W/m = 49.334 Btu/lb

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Explanation:

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5 0

4 years ago