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6

Essentially, yes. The derivation of the escape velocity is based only on the energy balance and energy does not depend on the direction. This follows from the property of the gravitational field to be conservative, so work required to move between any 2 points is independent from the path you take. Vague intuition: the vertical speed you estimate is about $...


5

Newton's third law tells us that the momentum imparted on one body is equal and opposite to the momentum imparted on another if they interact. We then have $$ \Delta \vec p_1~=~-\Delta\vec p_2. $$ The change in momentum is $\Delta \vec p_i~=~m\vec a_i\Delta t$, $i~=~1,~2$. The change in momentum is with Newton's second law due to a force so that $$ \vec F_1~...


4

In newtonian mechanics the angle does not matter, but in relativity it does. For example: An object close to the speed of light launched horizontally will orbit circular at a distance of 3GM/c² from the center of mass (the so called photon sphere), but it will escape if launched vertically. When you launch it at a distance just above 2GM/c² (the so called ...


3

The classical interference pattern is explained by the equations governing the behavior of light, and energy there is treated as a collective phenomenon, using the Pointing vector Energy transfer in a light beam can be best understood as an emergent phenomenon from the underlying quantum mechanical level. Innumerable photons create the visible interference ...


2

Newton's second law, force f is $$f=m\frac{d^2 x}{d t^2}$$ x is position vector of the particle. $$f=-\frac{d v}{dx}$$v is the potential energy. $$m\frac{d^2 x}{d t^2}=-\frac{d v}{dx}$$ Multiply both sides with $\dot x$ $$\frac{m}{2} \frac{d\dot x^2}{dt}=-\frac{dv}{dt}$$ $$ \frac{d}{dt}(\frac{1}{2}m\dot x^2+v)=0$$ i.e., $$\frac{dE}{dt}=0$$Energy is ...


2

If you accept that no external work was done, then if there is a change in the state of a system through which the kinetic energy changed, there must be a corresponding change in potential energy. The key to understanding the (rather poorly narrated) video is that the lecturer implies (at T=2:30) that $\Delta E=0$ from which it follows that $\Delta KE= - \...


1

A simple pendulum performs simple harmonic motion when it is displaced very slightly. You can say that a simple pendulum performs a periodic motion which can be treated as simple harmonic motion in small oscillation Now lets move forward supposing it to be pure SHM. The time period of a simple pendulum does not depend on how much it is displaced( but it ...



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