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The Quora quote is not useful because it does not tell you how to do any calculations. You need websites which explain how to calculate radius and period of orbit. eg http://www.physicsclassroom.com/calcpad/circgrav http://www.physicsclassroom.com/class/circles/Lesson-4/Mathematics-of-Satellite-Motion A more advanced webpage for calculating general (non-...


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Because we usually deal with continuous functions in physics. For instance, we know that internal energy of a body is continuous. I.e. if a body has internal energy of $U_1$ and $U_2$ in two different states and $U_2\gt U_1$ then certainly there are infinite states between 1 and 2 so that internal energy of the body is between $U_1$ and $U_2$. In other words,...


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We don't always need calculus to obtain a general form: static problems (a weight sitting stationary on a table, for instance) often don't require calculus. Calculus is therefore typically necessary where change occurs. Let's say a physical, observable quantity $u$ (distance, pressure or whatnot) changes in function of an observable parameter $t$ (time, ...


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We usually know the expression or "formula" for a small part, and this formula takes different values across parts, so one easy way to solve the problem, if this is the case, is by adding together all the mini formulas which is what an integral does. For instance you want to calculate the area inside a circle of radius R. How do you start? Well, if you ...


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This combination represents the following. $f$ in your reference is the number of particles per cubic space volume per cubic velocity volume. When you multiply it by some volume in velocity space, you obtain the regular density — number of particles per volume. But, as $f$ itself depends on velocity, to obtain the total density of all particles with all ...


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For the purposes of dimensions (units), you can treat a derivative like a division. So when you apply $\frac{{\rm d}}{{\rm d}t}$ to a function you divide the dimensions of the function by a unit of time. In your example I get: $$\frac{{\rm d}S}{{\rm d}t}\left[{\rm m}\,{\rm s}^{-1}\right] = v \left[{\rm m}\,{\rm s}^{-1}\right] + a\,\left[{\rm m}\,{\rm s}^{-2}...


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$\Delta x =v_{average}\times t$ In uniform acceleration $v_{average}$ becomes $\dfrac{v-v_{0}}{2}$ Hence; $\Delta x =\dfrac{v-v_{0}}{2}t=\dfrac{1}{2}\dfrac{v-v_{0}}{t}t^{2}=\dfrac{1}{2}at^{2}$


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This is a trivial kinematic deduction. \begin{align}s(t_2) &=\int_{t_1}^{t_2}~v(t)~\mathrm dt +s(t_1)\\ &= \int_{t_1}^{t_2}~\left\{v(t_1)+\int_{t_1}^{t}~a(t')~\mathrm dt'\right\}\mathrm dt+ s(t_1) \;.\end{align} Integrating this, we would get $$s(t_2)~=~ s(t_1) + v(t_1)\{t_2-t_1\} + a(t_1)\frac{\{t_2-t_1\}^2}{2} + \dot a(t_1)\frac{\{t_2-t_1\}^3}{6}...


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Indeed your first suggestion is wrong :$ \Delta x = v_o t + gt $ Instead it should be $ \Delta x = v_o t + gt^{2} $(You can recheck it) Where you are wrong is here: According to your question v is the final velocity since $(v=v_{0}+gt)$ So $\Delta x\neq vt$ but instead it should be $\Delta x =v_{average}\times t$ In uniform acceleration $v_{average}$ ...


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Let's take the first equation of motion which is : \begin{equation} v=u+at \end{equation} Integrate this equation to get: \begin{equation} \int\frac{dx}{dt}dt=\int{u}dt+\int{at} dt \end{equation} this gives: \begin{equation} x=ut+\frac{1}{2}at^2+x_0 \end{equation} The integration constant can be done away by putting the proper limits on $x$.(Assuming the ...


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I don't have a conceptual answer as to why it is that way. But mathematically, your first suggestion is wrong: $ \Delta x = v_o t + gt $. A unit analysis will show you why: $$ meters = \frac{meters}{seconds} seconds + \frac{meters}{seconds^2}seconds$$ $$ meters \not= meters + \frac{meters}{seconds}$$ And we can see that the assumption is simply not true ...


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There is no relation between the area being two-dimensional in your graph and what it means. For example, consider you make the $x$-axis as an indicator of the temperature, so what is the difference between $x= 5 K$ and $x = 8 K$? of course it's $ \Delta x = 3K $. Now, isn't $x$ a one-dimensional quantity? So the graph gains its meaning from you not from ...


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I'm confused that area is a 2 dimensional concept and it may indicate distance or displacement , which are 1 dimensional quantities. You are right. Area is a 2 dimensional quantity but what you have missed out is that you didn't use dimensional analysis properly. The dimensions of velocity is $[LT^{-1}]$. So if you multiply velocity with time, you get the ...


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If we consider the velocity-time graph area under the whole line is the distance. For example, Now look at the second part of the diagram (rectangular, dark blue) $$v=\frac{dx}{dt}\rightarrow\int{vdt}=\int{dx}\rightarrow x=vt$$ But the first part (triangle), we should consider acceleration $$a=\frac{d^2x}{d^2t}=\frac{dv}{dt}\rightarrow\int{adt}=\int{dv}\...


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The interpretation of the area under a curve, depending upon the curve, will vary. If it is a Velocity v. Time Graph, the area from a given time to another time, will be the distance traveled between those times. If it is an Acceleration v. Time Graph, the area from one time to another, will be the change in velocity of the object between those two times. ...


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The dimensions of the integral are simply those of $f(x)dx$, so in this case they would be $m^2/s^3 \times m = m^3/s^3$.


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It will be the latter case, $m^2/s^3m$ which is just $m^3/s^3$. Remember that the integral is the sum of all the products $f(x)\;\text{ times } \;dx$. $dx$ is a tiny piece of the path from $0$ to $x$, so it is in units of $m$ as well. Each of the products $f(x)dx$ have units $m^3/s^3$, and the sum of all these products keeps those units.


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Perhaps this is an easy way to see it. To first order in $h$, any tensor $T^{\mu \nu}$ can be written $T^{\mu \nu} = t^{\mu \nu} + \lambda h^{\mu \nu}$ for some $\lambda$ and tensor $\boldsymbol{t}$. Now, for the metric tensor, we must have that $g^{\mu \nu} = \eta^{\mu \nu} + \lambda h^{\mu \nu}$ by matching with the $\mathcal{O}(0)$ term, i.e. $g^{\mu \nu}...


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Whenever you see a function that looks like: $$ y = \tfrac{1}{2}kx^2 $$ there's a good chance it came from integrating the function: $$ \frac{dy}{dx} = kx $$ For example your distance function comes from integrating the velocity $v = at$: $$ y(t) = \int v\,dt = \int at\,dt = \tfrac{1}{2}at^2 $$ The spring energy function comes from integrating the ...


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Physicists do you the rules of calculus but can be sloppy in their notation. The symbols $\delta, \Delta$ and $d$ often seem to be used interchangeably to mean a (small or infinitesimal) change in something or better still a final value minus an initial value. So in your equation $dU$ is the change in internal energy of a system or final internal energy ...


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No you can't do that. The unit vector $\hat{a}_{\rho}$ does not mean: $\rho=1$, $\phi=0$, $z=0$. $\hat{a}_{\rho}$ actually depends on $\phi$ as it's $\left( cos\phi,sin\phi,0 \right)$.


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$$ a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = \frac{dv}{dx} v $$ The physics notation makes the truth of the derivation above clear because $dx/dx$ cancels. In mathematicians-style presentation of the rule, it's the chain rule for the derivative of a composite function, $v'(t(x)) =\dots$


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Treating $dx$ as a quantity you can multiply/divide both sides by is really just a shorthand for the chain rule, or equivalently u-substitution. Here's what's going on. You have $$ v=\frac{dx}{dt} $$ Thus, if you integrate both sides against time, they should (up to a constant) be equal $$ \int v dt = \int \frac{dx}{dt}dt $$ Doing a u-substitution on ...


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The term ${\rm d}t$ means an infinitesimal quantity of time. Like a slice of time so small, but still not zero. When something is moving with speed $v$ the infinitesimal distance it travels during the infitesimal time is designated ${\rm d}x = v \, {\rm d}t$. Speed just linearly scales time into distance. Now to get the total time passed, you sum of all ...


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v=dx/dt. It simply means that v is the slope of the x-t graph. v=dx/dt. dx=vdt. This is simple cross multiplication. Although I am not too sure, v=dx/dt has anything to do with deriving the equations of motion as it is acceleration that is the main factor in the equations of motion not velocity.



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