You drop an egg from rest with no air resistance. which of the following statements is correct? group of answer choices as it falls, its mechanical energy is not conserved, but its momentum is conserved. as it falls, its kinetic energy is conserved, but its momentum is not conserved. as it falls, neither its mechanical energy nor its momentum is conserved. as it falls, both its mechanical energy and its momentum are conserved. as it falls, its mechanical energy is conserved, but its momentum is not conserved.

As it falls, its mechanical energy is conserved, but its momentum is not conserved.

Explanation:

The mechanical energy of the egg during the fall is the sum of its potential energy and its kinetic energy:

E = U + K

As we know, mechanical energy is always conserved when there are only conservative forces acting on a system. In this case, the only force acting on the egg is the force of gravity, which is a conservative force (we are neglecting air resistance), therefore mechanical energy is conserved.

In particular, as the egg falls down, its gravitational potential energy U decreases, while its kinetic energy K increases (since its speed increases), but the sum of the two quantities, E, remains constant.

On the contrary, momentum is not conserved. In fact, momentum is given by

p = mv

where m is the mass and v is the velocity. Since the velocity of the egg increases (in magnitude) while it falls down, it means that momentum is not constant, so it is not conserved.

Why? The reason is the following: momentum is only conserved when the net external force acting on the system is zero. This is not the case: in fact, there is a net external force acting on the egg (the force of gravity), so the momentum cannot be conserved.

i'm on my way to history not a mistory.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft