<span>Direct Free Kick
Pass
<span>and Dribbling</span></span>
Answer:
B. Decrease the plane’s speed and height
Explanation:
Decrease in speed of plane:
When the speed of the plane is decreased then initial velocity of the package will be lesser. As a result, final velocity will also be lesser because in this case final velocity depends upon the initial velocity.
Decrease in height:
According to law of conservation of energy:
Loss in potential energy = Gain in Kinetic energy
mgh = ½ mV^2
It means with the decrease in height less potential energy will be converted to kinetic energy. As package has constant mass so it will gain less velocity.
It has a crystal stucture
Apply conservation of angular momentum:
L = Iw = const.
L = angular momentum, I = moment of inertia, w = angular velocity, L must stay constant.
L must stay the same before and after the professor brings the dumbbells closer to himself.
His initial angular velocity is 2π radians divided by 2.0 seconds, or π rad/s. His initial moment of inertia is 3.0kg•m^2
His final moment of inertia is 2.2kg•m^2.
Calculate the initial angular velocity:
L = 3.0π
Final angular velocity:
L = 2.2w
Set the initial and final angular momentum equal to each other and solve for the final angular velocity w:
3.0π = 2.2w
w = 1.4π rad/s
The rotational energy is given by:
KE = 0.5Iw^2
Initial rotational energy:
KE = 0.5(3.0)(π)^2 = 14.8J
Final rotational energy:
KE = 0.5(2.2)(1.4)^2 = 21.3J
There is an increase in rotational energy. Where did this energy come from? It came from changing the moment of inertia. The professor had to exert a radially inward force to pull in the dumbbells, doing work that increases his rotational energy.
To solve this problem it is necessary to apply the concepts related to the Heisenberg's uncertainty principle. Under this principle we understand the relationship that the minimum range of error in position (x) times the minimum range of error in momentum (p) is, at a minimum, about equal to the Planck constant, mathematically that is,

Replacing with our values we have,


Therefore the least uncertainty in any simultaneous measurement of the momentum component px of this electron is 