energy of a partical in free fall at different position of partical and explain conservation ij free fall
Answers
Answer:
Douglas-fir, spruce, true fir, beech, and maple are toward the top of the list for oxygen release.Lysosomes pass through various stages in the same cell. The phenomenon is called polymorphism or existence of more than one morphological form. Depending upon their morphology and function, there are four types of lysosomes— primary, secondary, residual bodies and auto-phagic vacuoles (Fig. 8.33).Lysosomes pass through various stages in the same cell. The phenomenon is called polymorphism or existence of more than one morphological form. Depending upon their morphology and function, there are four types of lysosomes— primary, secondary, residual bodies and auto-phagic vacuoles (Fig. 8.33).Ways people develop anosmia include:
Head injuries.
Severe injuries to the upper part of the nose.
Severe upper respiratory infections.
Polyps in the sinus cavities or nasal passages.
Tobacco smoke.
Inhaling certain chemicals such as pesticides or ammonia for long periods of time (years)
Hormonal imbalances.
Answer:
Consider a mass $m$ which is falling vertically under the influence of gravity. We already know how to analyze the motion of such a mass. Let us employ this knowledge to search for an expression for the conserved energy during this process. (N.B., This is clearly an example of a closed system, involving only the mass and the gravitational field.) The physics of free-fall under gravity is summarized by the three equations (24)-(26). Let us examine the last of these equations: $v^2 = v_0^{ 2} - 2 g s$. Suppose that the mass falls from height $h_1$ to $h_2$, its initial velocity is $v_1$, and its final velocity is $v_2$. It follows that the net vertical displacement of the mass is $s=h_2-h_1$. Moreover, $v_0=v_1$ and $v=v_2$. Hence, the previous expression can be rearranged to give
\begin{displaymath}
\frac{1}{2} m v_1^{ 2} + m g h_1 = \frac{1}{2} m v_2^{ 2} + m g h_2.
\end{displaymath} (123)
The above equation clearly represents a conservation law, of some description, since the left-hand side only contains quantities evaluated at the initial height, whereas the right-hand side only contains quantities evaluated at the final height. In order to clarify the meaning of Eq. (123), let us define the kinetic energy of the mass,
\begin{displaymath}
K = \frac{1}{2} m v^2,
\end{displaymath} (124)
and the gravitational potential energy of the mass,
\begin{displaymath}
U = m g h.
\end{displaymath} (125)
Note that kinetic energy represents energy the mass possesses by virtue of its motion. Likewise, potential energy represents energy the mass possesses by virtue of its position. It follows that
E = K + U = Constant.