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1

Events can 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.


0

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 ...


0

We can draw a tangent (Call it 't') at any point (call it 'p')on a curve. We can also draw a circle to which 't' is a tangent. The radius of this circle is the radius of curvature to the given curve at the point 'p'. An analogy from motion of a body along a curved path may help easier understanding. When a body moves along a curved path, its velocity keeps ...


0

So the x part of the drag is $$\begin{align*} f_x &= f \cos\theta \\ &= (k v^2) \cos\theta \\ &= k v (v \cos\theta) \\ &= k v v_x \,, \end{align*}$$ and $v$ is dependent on the $y$ component of motion as well as the x-component. Similar consideration, of course, apply to the y-component of drag. So the independence is explicitly lost and ...


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

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 ...


0

The situation is quite possible.As you have not mentioned any Exact value of acceleration or time between the two lights opening it is difficult to tell exact values which will satisfy your event.But I have made a similar thing which may help you understand. Imagine AE is the train and B,C,D are points on it such that AB=BC=CD=DE.Now AE is moving with ...


0

Your entire analysis is entirely correct and complete (and in my opinion the best way to go about it). There's no more information that need to be derived about this issue. Other analyses are just a different representations of the same facts.


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 ...


0

Hint: Calculate the vertical velocity as a function of time ($v_y=-g\cdot t$). Two vectors are perpendicular when their dot product is zero, so $$v_{x1}\cdot v_{x2} + v_{y1}\cdot v_{y2}=0$$ Now $v_{y1} = v_{y2} = gt$ since both are free falling. So you find that $gt = \sqrt{12}$ when they are perpendicular. From this, you calculate the time; and from the ...


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.


1

As long as your units are consistent you can use any units you want and the equation will be unchanged. A change would only be required if you started mixing up the units e.g. specifying the velocity in m/s and the acceleration in km/h$^2$.


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 ...


0

D=vt is used when acceleration is zero but D=ut+1/2 at^2 is used when there is a constant acceleration.


0

The velocity is always, by definition, the derivative of the position $\textbf{r}(t)$ with respect to the time and likewise for the acceleration $$ \textbf{v}(t)=\frac{d}{dt}\textbf{r}(t),\qquad \textbf{a}(t)=\frac{d}{dt}\textbf{v}(t)=\frac{d^2}{dt^2}\textbf{r}(t). $$ The position is, in turn, always a function of the time, although often not explicitly ...


0

Acceleration is defined as the derivative of velocity with respect to $t$: $$a=\frac{dv}{dt}$$ It is the instantaneous change of velocity. Just like velocity is defined as the instantaneous change of position $r$: $$v=\frac{dr}{dt}$$ If you agree that: $$a=-\frac{GM}{r^2}$$ then it is a simple thing to exchange $a$ with its definition $dv/dt$.


0

Because by definition acceleration is change in velocity over time. The fact that acceleration is a function of radius changes nothing. Basically what the equation says is that acceleration as a function of radius is equal to some formula. We then know from the definition of acceleration that it is change in velocity over time. If this helps clear it up: ...


0

A parabola does not have a radius of curvature. Each point of the parabola does. So, if you are asked to find the radius of curvature on a parabola, you must also be give the point on the parabola.


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} ...


0

is $\left|\frac{d\vec{r}}{dt}\right| = \frac{d|\vec{r}|}{dt} $ [?] There's a subtlety here ... It depends on how (you want that) the right-hand side of your suggested equation is interpreted. We know, of course, by to the rigorous definition of what the notation you've used is supposed to mean, that the value of a derivative is evaluated "at" a ...


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 ...


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

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 ...


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

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

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 $f(t)$ ...


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 ...


0

A certain minimum force is needed to move the wheels. If you move your body slowly, you can move forwards (or back) on the chair without the chair moving. But if you make a sudden move, the force you exert on the chair (and that the chair exerts on you) is greater than this minimum. This means you can move by using a different acceleration in the forward ...


0

If your object has mass $M$ and you want to accelerate it with acceleration $a$ to a specific end-velocity $v$ you have to keep in mind that the energy $$e = \frac{M\cdot v^2}{2}$$ and also $$e = F\cdot x$$ where $x$ is the distance over which the force $F$ which equals $M\cdot a$ is applied. Knowing that you can solve for the distance over which you ...


0

work - kinetic energy theorem $$ Fx = \frac{m}{2}(V_1^2 - V_0^2) $$ if acceleration is constant, $$ x = \frac{V_1 + V_0}{2}t $$


0

My thoughts on this problem: There are two variables, yes: $t$ and $v_{woman}$. Solve for t! You get 0, 1 or 2 solutions, depending on $v_{woman}$. The problem says, however, that you want the minimum $v_{woman}$ velocity. If you get 0 solutions, for t, it means that they'll never meet. If you get two, then the woman gets in front of the train, which after ...


0

The answers are exactly equivalent, the dot product between two vectors produces the product between magnitude of the vectors and the cosine of the angle between the vectors... In this case the angle between two c vectors will be 0 and the angle between two (bt) vectors will also be 0, hence the dot product will be equal to the square of the magnitude of ...


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

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 ...


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 ...


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 ...


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

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.


0

The energy difference between two arbitrary speeds $v$ and $v'$ cannot be found by $\frac{m(v'-v)^2}{2}$ except for the case where $v=0$. So your calculation about the energy change between $50m/s$ and $25m/s$ is incorrect. You instead have to calculate the initial and final energy states and subtract directly. $$E(50) = \frac{(20kg)(50m/s)^2}{2}$$ ...


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 ...


0

do it with conservation of energy first of all find out the potential energyat the top of the semi circular bowl keep it equal with the kinetic energy at lowest point


0

As CuriousOne wrote, the device will lack frequencies ( and power ) The general solution is to evaluate the efficiency of Time reversal signal processing. If the bell ringing is , each time it is processed , identical ( rarely true ) the aerology between the bell and you is each time identical you are able to record all the sound and to reproduce it ...


0

The question implies that acceleration is constant. Newton's second law states: $F=ma$ (for constant mass $m$) The fact that he asked for "the" force, implies that it's a constant force, and a constant force would result in constant acceleration of the body. the two above expressions you wrote for constant acceleration are equivalent: $a$=$\Delta ...


-2

"a = ΔV / ΔT" and "Vf = Vo + at" are the same really. ΔV is Vf - Vo, otherwise known as "Change in Velocity" So.. Vf = Vo + at at = Vf - Vo a = (Vf - Vo)/t a = ΔV/ΔT To calculate the Force required, a = ΔV/ΔT F = m*a or a = F/m F/m = ΔV/ΔT F = (ΔV*m)/ΔT


0

Actually the pitch obtained by striking a coffee cup is a function of how full the cup is with coffee - the fuller the cup, the lower the pitch. You can convince yourself of this with a simple experiment just by changing the amount of coffee (water will work) and tapping the outside of the cup with a spoon. But when you stir a cup of coffee with a fixed ...


0

The value of the time coordinate $t$ of an event, just like the value of its spatial coordinates $x,y,z$ is unimportant, totally arbitrary and differs from one observer to another. What does not change however,(at least in classical mechanics) is the spatial separation $\Delta x$ between two events and the time interval $\Delta t$ that separates them. By ...


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

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.


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 ...


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)



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