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Sindrei [870]
1 year ago
14

an electron and a 0.0320-kg bullet each have a velocity of magnitude 510 m/s, accurate to within 0.0100%. within what lower limi

t could we determine the position of each object along the direction of the velocity?electron mmbullet m
Physics
1 answer:
MArishka [77]1 year ago
6 0

for this we apply, Heisenberg's uncertainty principle.

it states that physical variables like position and momentum, can never simultaneously know both variables at the same moment.

the formula is,

Δp * Δx = h/4π

m(e).Δv * Δx = h/4π

by rearranging,

Δx = h / 4π * m(e).Δv

Δx = (6.63*10^-34) / 4 * 3.142 * 9.11*10^-31 * 5.10*10^-2

Δx = 6.63*10^-34 / 583.9 X 10 ⁻³¹

Δx = 0.011 X 10⁻³

for the bullet

Δx = (6.63*10^-34) / 4 * 3.142 * 0.032*10^-31 * 5.10*10^-2

Δx = 6.63*10^-34 /2.05

Δx =3.23 X 10⁻³² m

therefore, we can say that the lower limits are   0.011 X 10⁻³  m for the electron and  3.23 X 10⁻³² m   for the bullet

To know more about bullet problem,

brainly.com/question/21150302

#SPJ4

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Explanation :

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