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When you say "why aren't things being destroyed", you presumably mean "why aren't the chemical bonds that hold objects together being broken". Now, we can determine the energy it takes to break a bond - that's called the "bond energy". Let's take, for example, a carbon-carbon bond, since it's a common one in our bodies. The bond energy of a carbon-carbon ...


4

If the object was released from rest ($u=0\,{\rm m/s}$), what is its speed after $1\,{\rm s}$ if $a=10\,{\rm m/s}^2$? Using: $$v=u+at$$ you will find that the object was not released from rest...


2

You must solve for the objects initial velocity first: $$ v(t)=u+at\\ v(0)=u\\ v(1)=u+10\text{m/s}\\ =20\text{m/s}\\ u=10\text{m/s} $$ With this adjustment you should find the correct answer.


2

Things actually do get destroyed by what those air molecules pick up and throw around. Take look at this example [image from here: http://en.wikipedia.org/wiki/File:Arbol_de_Piedra.jpg ] Just like their bigger sized brothers, it's the load of those mini-torpedos that brings the destruction.


2

Ok, so here are is my solution. I'll be happy is someone can provide something simpler. $a_1, v_1, t_1 - acceleratin, terminal\ velocity,\ time\ to\ reach\ it\ (for\ Tiger)$ $a_2, v_2, t_2 - acceleratin, terminal\ velocity,\ time\ to\ reach\ it\ (for\ You)$ $s_0 - starting\ distance$ Not let's consider three cases: 1) $a_1 < a_2 \wedge v_1 < v_2$ ...


2

It's worth drawing a diagram: The equations of motion are: $$x_1 = x_0 + \frac12 a_1 t^2$$ for $t_1 < \frac{v_1}{ a_1}$ And $$x_1 = x_0 + \frac{v_1^2}{2 a_1} + v_1 (t - \frac{v_1}{a_1})$$ for the steady state. The same equations, with different suffixes, hold for the tiger. To solve, you start by considering all different orders of $t_1, t_2, ...


1

This can be understood in terms of vector differentiation and the dot product. Take the example that $v \perp r$. The change in the square of the displacement is $$\frac{d}{dt}r^2$$ $$=2r \cdot v$$, and if they are perpendicular, the dot product is zero.


1

Let's say you have 3 systems. $B$ moving relative to $C$ with velocity $u$ and $A$ moving relative to $C$ with velocity $v$, all along one axis. $A$ will "measure" for the velocity of $B$: $$ u' = \frac{u-v}{1-\frac{uv}{c^2}} $$ While $B$ will "measure" for the velocity of $A$: $$ v' = \frac{v-u}{1-\frac{uv}{c^2}} = -u' $$ It holds true that the velocity ...


1

The velocity addition formula you cite does not quantify what you think it quantifies. I'm going to say see and when I do, I mean "computes relative to its own frame". The velocity addition formula describes the following setup. Frame 0 sees a particle (particle 1) moving in a direction at speed $v_1$. Frame 1 sees that particle 1 at rest and frame 1 ...


1

The force due to sliding friction causes the center of mass to decelerate - since it is the only external force acting on the ring, and we know the center of mass moves as though all forces act there. You might find this earlier answer of mine and the links therein helpful. As for the "in the real world the ring will stop completely" part of your question - ...



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