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Oxana [17]
3 years ago
9

A particle is confined to a one-dimensional box that is 50 pm long. What is the smallest possible uncertainty in momentum for th

e particle?
Physics
1 answer:
nevsk [136]3 years ago
6 0

Answer:

The smallest possibility is 0.01E-22kgm/s

Explanation:

Using

Momentum= h/4πx

= 6.6x 10^-34Js/ 4(3.142* 50*10-12m)

= 0.01*10^-22kgm/s

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The vertical component of the force is 100 N.

The given parameters:

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<h3>Horizontal component of the force</h3>

The horizontal component of the force is calculated as follows;

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The vertical component of the force is calculated as follows;

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An airplane flying at a speed of 130 mi/h (58.1 m/s) and at an altitude of 4.9 km drops a food package. Without a parachute, at
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The food package will strike the ground at 11 degrees below the horizontal.

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The time for the food package to hit the ground is calculated as follows;

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vf(y) = vo + gt

vf(y) = 0 + (31.62 x 9.8)

vf(y) = 309.88 m/s

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The horizontal component of the speed will be constant, while vertical component will change

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