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castortr0y [4]
3 years ago
13

Consider an electron confined in a region of nuclear dimensions (about 5 fm). Find its minimumpossible kinetic energy in MeV. Tr

eat this problem as one-dimensional, and use the relativistic relationbetweenEandp. Give your answer to 2 significant figures. (The large value you will find is a strongargument against the presence of electrons inside nuclei, since no known mechanism could contain anelectron with this much energy.)
Physics
1 answer:
BARSIC [14]3 years ago
7 0

Answer:

39.40 MeV

Explanation:

<u>Determine the minimum possible Kinetic energy </u>

width of region = 5 fm

From Heisenberg's uncertainty relation below

ΔxΔp ≥ h/2 , where : 2Δx = 5fm ,  Δpc = hc/2Δx = 39.4 MeV

when we apply this values using the relativistic energy-momentum relation

E^2 = ( mc^2)^2 + ( pc )^2 = 39.4 MeV ( right answer ) because the energy grows quadratically in nonrelativistic approximation,

Also in a nuclear confinement ( E, P >> mc )

while The large value will portray a Non-relativistic limit  as calculated below

K = h^2 / 2ma^2 = 1.52 GeV

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Certain neutron stars (extremely dense stars) are believed to be rotating at about 1000 rev/s. If such a star has a radius of 14
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3 years ago
Suppose you lift a 20 kg box by a height of 1.0 m.
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If the ball is 0.60 mm from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a
PolarNik [594]

This question is incomplete, the complete question is;

In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.

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Angular Velocity at time 0.15s = 24 rad/s

a) What is the angular acceleration?

b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g

Answer:

a) the angular acceleration is 80 rad/s²

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Explanation:

Given the data in the question;

from the graph below;

Angular Velocity at time 0s w_o = 12 rad/s

Angular Velocity at time 0.15s w_f = 24 rad/s

a) What is the angular acceleration;

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Angular acceleration ∝ = ( 24 - 12 ) / 0.15

Angular acceleration ∝ = 12 / 0.15

Angular acceleration ∝ = 80 rad/s²

Therefore, the angular acceleration is 80 rad/s²

b)

If the ball is 0.60 m from her shoulder, i.e s = 0.6 m

the tangential acceleration of the ball will be;

a = ∝ × s

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a = 80 × 0.6

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Answer:

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