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22

So let's start with your last question, informally, the radius of curvature is a measure of how much a certain curve is pointy and has sharp corners. Given a curve $y$, you can calculate its radius of curvature using this formula: $$\dfrac{\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^\dfrac{3}{2}}{\left|\dfrac{d^2y}{dx^2}\right|}$$ You might ask what ...


3

Actually you're asking two different questions. Is the magnitude of instantaneous velocity the same as instantaneous speed? Well, yes, that's the definition of instantaneous speed. Is this equation true? $$\biggl\lvert\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\biggr\rvert = \frac{\mathrm{d}\lvert\vec{r}\rvert}{\mathrm{d}t}$$ No, it's not - but instantaneous ...


3

Your equation is not valid, see Figure below With equations \begin{equation} \Vert\mathbf{r}\Vert^{2}=\mathbf{r}\circ \mathbf{r} \Longrightarrow 2\cdot \Vert\mathbf{r}\Vert\cdot d\Vert\mathbf{r}\Vert =2\cdot\left(\mathbf{r}\circ d\mathbf{r}\right)\Longrightarrow d\Vert \mathbf{r}\Vert =\dfrac{\mathbf{r}}{\Vert\mathbf{r}\Vert}\circ d \mathbf{r} \tag{01} ...


2

Yes, if you use same units then there will be no change. If you use all measurements in fps then also formula remains valid. It is universal, given all units are in same scale.


2

The answer by Bryson S. is solid, thorough and vey good, as is Jerry Schirmer's hint. This is merely another way of looking at the problem. We can consider, as Jerry Schirmer points out, two components of acceleration; a tangential and a normal component. Before we begin, note that velocity always points tangential to the path a particle travels. This is ...


2

The answers are actually equivalent. $|\vec{c}|^2 = \vec{c} \cdot \vec{c} = \Sigma_i x_i^2$ Where the $i$'s run over whatever number of dimensions you have. So you're both right.


2

In general, assume you have some time-varying quantity $f(t)$, that varies from some time $t_{0}$ to some other time $t_{f}$. Now, graph $f(t)$. There will be some area under this curve. The average value of $f(t)$ over this time interval, $\bar f$ will be the height of the square that has the same base as the graph of $f(t)$ and the same area as $f(t)$ ...


2

Well, this subject deserves some effort and a help form philosophy of science. But if you are looking for a mathematical and physical approach I would suggest you to read the paper by Carlo Rovelli. Here's the link. Summarizing, the paper argues that within the conditions Aristotle lived his physics can be taken as an empirically grounded theory.


2

Equation (1) is just $\ddot x=g\sin\theta$ multiplied by $2\dot x$ on both sides. If you understand $\ddot x=g\sin\theta$, this should follow directly. For your second question, we use the chain rule, $$\frac{d}{dt}\dot x^2=\frac{d\dot x^2}{d\dot x}\frac{d\dot x}{dt}=2\dot x\ddot x.$$


2

First of all, derivative of $1.5t-9.75t^3$ is $1.5-29.25t^2$, you missed first part. Secondly, consider ideal pendulum. It's speed is zero at extreme points, however it never stops for ever. it depends on the problem and additional conditions (like, object hitting ground, so speed=0 is good enough condition for finding time of impact)


2

I would go for this: Imagine the bottom of the cup as a saw. The noise or chattering of the spoon jumping on the sawteeth is higher the faster spoon moves. Those "sawteeth" on the cup bottom are very small, but the principle is the same. Therefore the faster stirring the higher pitch.


2

You can easily get that answer by noticing that in when you have constant acceleration the average velocity after a time $t$ is: \begin{equation} \langle v\rangle_{acc.=w}=\frac{v_{start}+(v_{start}+w t)}{2} \end{equation} for your case $v_{start}=0$ and then $\langle v\rangle_{acc.=w}=\frac{wt}{2}$. For the deceleration we get the same result. Now you ...


2

You first write your position vector as $\vec r = (x_0 -t\cdot10~\mathrm{m/s},y_0)$ and then take the derivative of that. This produces $$\vec v = \frac{d\vec r}{dt}=(-10,0)~\mathrm{ms^{-1}}=-10\hat i\,~\mathrm{ms^{-1}}$$ which is what you want.


2

$\newcommand{\dv}[2]{\frac{\mathrm{d} #1}{\mathrm{d}#2}}$Elaborating on Mikael's answer, note that equations like $\vec v = \dv{\vec r}{t}$ are sort of shorthand notations to make the life of a physicist easier. Note that there are two (three in 3d) equations in this condensed notation. What we mean by such equations is simply the following: $$v_x = ...


2

If you have the position vector along a path $\vec{r}(q)$ parametrized by $q$, where $q$ can be time, angle, distance, or whatever then the derivatives are: $$ \vec{v}(q,\dot{q}) = \frac{\partial \vec{r}(q)}{\partial q} \dot{q} $$ $$ v = \| \vec{v} \|$$ $$ \vec{e} = \frac{ \vec{v}}{v} $$ $$ \vec{a}(q,\dot{q},\ddot{q}) = \frac{\partial ...


1

The formula for kinetic energy is $\frac12mv^2$. If your initial velocity is $v_i$ and your final velocity is $v_f$, then your initial kinetic energy is $KE_i = \frac12 m v_i^2$ and your final kinetic energy is $KE_f = \frac12 mv_f^2$. The difference is $\Delta KE = KE_f - KE_i = \frac12 m(v_f^2 - v_i^2)$ It appears you're thinking that you can define ...


1

The acceleration is the time derivative of the velocity: $$ a = \frac{dv}{dt} $$ so if the velocity does not change with time the acceleration is necessarily zero. Since in your example the velocity is constant during the interval that means $dv/dt = 0$ and therefore that $a = 0$ during the interval. The velocity doesn't have to be zero. Any constant ...


1

This is my solution. Place $m$ to the left described by the coordinate $x_1$ and $M$ to the right, described by the coordinate $x_2$, both coordinates increasing to the right. Take $\vec{F}$ to be acting upon $M$ to the right and $\eta_0$ to be the natural separation of the spring. Finally, take $x$ to be the center of mass of the system and $\eta$ be the ...


1

In polar coordinates you have $(x,y) = (r \cos \theta, r \sin \theta)$ Taking total derivatives of the above one finds that: Positions $$\begin{aligned} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{vmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix} \begin{pmatrix} r \\ 0 \end{pmatrix} \end{aligned} $$ Velocities ...


1

Mythbusters ran an episode on this one. When you learn of relative velocity, you learn that from the perspective of one of the drivers, the other driver will appear to come at you at 80 m/s. So the thought is that it should be the same as hitting a wall at 80 m/s. The energy doesn't add up though. Essentially, from the perspective of the driver, they are ...


1

Obviously not: think to a very simple (2d) example, $r(t)=(t,t)$. Componentwise, the derivative yields $1,1$, and hence $\lvert dr/dt\rvert=\sqrt{2}$. On the other hand, $\lvert r(t)\rvert = \sqrt{2}\lvert t \rvert$. And the absolute value function is not differentiable in zero. Hence the two derivative functions coincide almost everywhere, but not in ...


1

The total kinetic power of the system will be $\frac{1}{2}mv^2_1 +\frac{1}{2}mv^2_2$. The first equation that you mention is wrong, because this equation says that you have an object of mass $m$ with speed $v_1+v_2$. If you expand the squared term you will see that it is different. Now, what do you mean hit each other? Do they have opposite velocities? In ...


1

First of all, as already mentioned, yes $t$ is in units of hours. An equility remains an equality after introducing units. Hence, the unit on the RHS and the LHS have to agree and in fact also the sum of two quantities (as $3.6m/s+ 1km/h$) can only be simplified when both have the same unit! Clarifications of the appearance of units Units are actually a ...


1

The answer is, "as long as the light from the front of the train has not hit the back of the train first, then yes, there exists some reference frame which thinks of both as simultaneous." In special relativity, there is a number which everyone agrees on: take any two events that are separated by a vector (in your coordinates) $\vec r$ and time interval ...


1

I am going to assume that you have not yet studied linear algebra, sorry if it seems as if I am talking down to you at any point. You are correct in that we can split a vector into two components in the plane. This is because any two linearly independent(not parallel or anti-parallel) vectors form a basis(a set of vectors from which you can "build" other ...


1

Let me describe a geometrical way to approach your question. We start by drawing a spacetime diagram with distance along the horizontal axis and time on the vertical axis. We'll also take the speed of light to be $1$, so on our diagram a light ray travels along the line $x = t$ i.e. at 45º. If we have some other observer moving at a speed $v$ relative to ...


1

Events can never univerally only be said to happen in a particular order if they are space-like events, i.e. $(\Delta r)^2 > c^2(\Delta t)^2$. If this is true, the there is no universal agreement about the order in which the events happen. If, however, $(\Delta r)^2 ≤ c^2(\Delta t)^2$ holds true, then every observer will agree on the order.


1

You conclude that constant acceleration is appropriate for one of a few reasons: You read in the question text that you should treat it that way. If (1) does not apply, you read in the question text that some physical situation obtains and you know or suspect that this situation is usually well represented by a constant acceleration. If neither (1) nor (2) ...


1

I believe you are correct to use the first two equations you listed. The velocity of the cart going into the turn is (as you stated) $v = \sqrt{ 2gh }$ because the potential energy $U = mgh$ is transformed completely into kinetic energy $KE = \frac12 mv^2$ assuming the loop is on the ground (where $U=0$) and energy lost to friction is also 0. Now that the ...


1

Your last equation is a quadratic in $t$. The $a$ is simply $\sin(\theta).g$. You can then solve it with the usual formula for a quadratic equation. There are two solutions to a quadratic, and that's because if you go into negative time you'd be pulled down by gravity any get to the new X position. This solution, of course, wouldn't apply to your ...



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