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0

Yes you would actualy see the car with its speed added with your speed.


1

Your understanding is correct. And switching between different perspectives (what physicists would call different inertial frames of reference) like that is a very useful tool in physics, because it turns out that the laws of physics have the same form no matter which inertial frame of reference a problem is described in. For more information, see ...


2

Rough impressions can be misleading. The other car really is moving 200 km/h from your point of view. One thing to keep in mind is that you tend to perceive motion more readily when it is closer to you. This is at least partly due to the fact that you really only can see angular speed across your field of view (like degrees per second). To convert this into ...


1

when low mass object hits high mass object it is reflected gaining opposite velocity almost the same as initial velocity. If I jump onto the wall why my body is not reflected? I know that collision is not fully elastic but it should be at least similar. Human body is not elastic: it cannot be deformed/ compressed in any way and then return to ...


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Changing shape could still be an elastic deformation (of a rigid body). So obviously there are also plastic deformations involved, when jumping against a wall.


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WE know that the equation of motion of a simple pendulum can be written as y= a sin(wt). y= displacement of the pendulum at time "t" from the mean position a= amplitude of oscillation w= angular frequency of oscillation remember in this case y=0, when time t=0. now (dy/dt)= aw cos(wt)= velocity =v again (dv/dt) = ...


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I'd like to add something to these answers. In the classical mechanics, we cannot distinguish a moving body from the body at rest, if we look at it at any particular instant. So, we have to add some hidden information to the picture, that is instantaneous velocity. But that's what physics only knew in the 19th century. In 20th century physics, there have ...


4

At a "frozen" instant of time, the arrow may not be moving - but this is a tautology, since movement is something that requires time. However, even in that frozen instant the arrow does have a velocity (instantaneous velocity, if you will). Imagine that time is a series of huge number of discrete frames (or instead imagine that it is continuous, and that we ...


1

Even though the forces started at different times, is there any displacement of the metal box in any of the situations? Or is there any movement at all but is the net displacement zero? Sure. If you think of each force as causing an acceleration, the first one begins an acceleration in one direction, the second an acceleration in the other (or a ...


1

"Most probably in reality there are some extremely complex laws and equations which makes this question more complicated." Not really. The equations are rather straightforward. Let's measure velocities in units of the speed of light and let's denote the velocity of $B$ as observed by $A$ as $v_{BA}$, the velocity of $A$ as observed by $C$ (the ...


1

That light has a fixed velocity in vacuum comes from observations . In order to fit the data Lorenz transformation were imposed on the rigorous mathematical model for electromagnetism, Maxwell's equations. It was the result of attempts by Lorentz and others to explain how the speed of light was observed to be independent of the reference frame, and to ...


4

In your frame of reference, it does indeed look as though the difference in speed between A and B is greater than $c$. But the question is - does A think that B is moving away at that speed? And the answer is "no". There is a thing called the Lorentz transformation which describes how the observed speed of an object is a function of the speed of the ...


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I doubt this is your answer: what is the frequency $f$ doing, in the problem of a magnet sliding down an incline? What you should ask yourself is: how is changing the thickness of the aluminum foil going to modify the time down the incline? Increasing the thickness of the aluminum foil increases/decreases/leaves unaltered the foil's resistance (you pick ...


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Taking answers and comments into account, my own current conclusion is that velocity is an intensive property, provided the system considered is homogeneous, at least with respect to speed. Like other intensive properties, this may depend on scale, and cease to have meaning at molecular level. I did not intend to write an answer to my own question but ... ...


3

Your question, as of right now, seems confused to me. An extensive property of a system is one that scales with the system size. An intensive property is independent of the system size. For example, consider a system $A_1$ with $N$ particles in a volume $V$, with density $\rho=\frac{N}{V}$. Now, we consider two of these systems separately, $A_1$ and $A_2$, ...


3

why does it seem improper to add many speeds (or velocities)? Adding speeds is ofttimes inappropriate even in Newtonian mechanics. Suppose Mark is moving 3 m/s eastward with respect to Bob, and John is moving 3 m/s westward with respect to Mark. The relative velocity between Bob and John is zero rather than the 6 m/s suggested by adding speeds. You can ...


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Your first idea would work if it was possible to accelerate a rocket to the speed of light rather than only being able to approach the speed of light. But a person on a rocket moving at $.99c$ relative to Earth will still look out his window and see photons zipping past at the speed of light, not at $.01c$. Since it would require infinite energy supplied to ...


1

Where's the flaw? Here's one: Let γ be the squareroot of 1−v²/c² but, in fact, $$\gamma_v = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$ But, there is a flaw in your reasoning too so correcting the error in the formula for $\gamma$ will not get you to the correct answer. Let's work it out with coordinates to see explicitly what's going on. Let the ...


3

Without repeating the mathematics of @alemi's answer, let's just think about one other thing: Buoyancy is a function of depth - that is, as you descend your lungs are compressed and your buoyancy decreases, then goes negative (depending on the fat content of your body - if you have enough fat, you remain positively buoyant at any depth since you do not need ...


7

I like Jerry Schirmer's answer, but I was worried that instead of modelling you swimming as fast as you can as a constant force, I thought it would be interesting to consider swimming as fast as you can as a constant power. This seems more logical to me as if you are really trying to go as fast as you can, you will be limited by how hard you work, how fast ...


1

Since you know the size of the inlet pipe, find the area of the hole using $A=\pi r^2$. Multiple the area by 10. Since the drum is round, find the radius using $r=\sqrt {\frac {A}{\pi }}$. Multiply the radius by 2 to get the diameter.


2

OK, lets solve this in as simple an approximation as possible. I"m going to assume that the whole trip happens at the appropriate terminal velocity, and that acceleration times are very small. Furthermore, I"m going to model the resistant force${}^{1}$ of the water as $F = cv^{2}$, for some constant $c$. On the way down, you have the the resistant force ...


5

According to the RSB: Migrating (European) swallows cover 200 miles a day, mainly during daylight, at speeds of 17-22 miles per hour. The maximum flight speed is 35 mph I think that calculating the flight speed of a bird from first principles would be impossibly difficult as there are just too many variables. The web site you link does get an answer ...


0

From the path you need to find the radius of curvature $\rho$ at each point. This would be kind of noisy unless you have really precise data. Your best bet into input all the x and y points into cubic spline in order to get what the derivatives $x'$ and $y'$ are (in units of length per frame). In addition, you need to get the kinematic accelerations $x''$ ...


0

In your case, lets $\Delta t = 0.03s $ By the method alemi explained, $$a_{x}(t)=\frac{x(t-\Delta t)-2x(t)+x(t+\Delta t)}{(\Delta t)^{2}}$$ and $$a_{y}(t)=\frac{y(t-\Delta t)-2y(t)+y(t+\Delta t)}{(\Delta t)^{2}}$$


2

Yes. Provided you are only interested in the direction of the acceleration, and not it's magnitude. And further assuming your time samples are equally spaced, you can take the second derivative of the path and this will be proportional to the acceleration. A decent method in practice would be to use a second order central finite difference scheme wherein ...


5

That's a very hard question to answer theoretically because the aerodynamic drag, and therefore the terminal velocity, is highly dependant on the shape of the falling object. Assuming we're in the turbulent regime, the aerodynamic drag is given by the equation: $$ F_d = \tfrac{1}{2}\rho v^2 C_d A $$ where $\rho$ is the density of the air, $v$ is the ...


3

Like humans, the sheep has some – although more limited – freedom to act if it wants to change the asymptotic speed. A skydiver's asymptotic speed may be between 50 m/s in the "face down" free fall and 90 m/s when he or she pulls the limbs in. These two speeds hugely differ which indicates that it doesn't make much sense to talk about a "universal" value, ...


1

If I understand your code correctly, your total force $F$ is given by an variable throttle force $F_T$ and a drag term whose sign is the same as the sign of the velocity (i.e. the factor (m_speed > 0.0 ? 1.0 : -1.0), which I believe equals $\text{sgn}(v)$. There is a problem, I think, in the velocity dependence of that drag term. You are currently ...


1

No. According to special relativity's postulates, the speed of light in a vacuum is the same for all observers, regardless of the motion of the light source. You may want to read the relativistic Doppler effect. It refers to the change in frequency of light due to motion of observer and/or source.



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