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37

That's not true, Newtons's laws do not say that. What's important here is conservation of momentum. Inside the phone, there is an oscillating mass. While the mass inside has a momentum and thus a velocity in one direction, the (friction-free) phone has to have the same momentum in the opposite direction. It "vibrates". Homework: Get on a skateboard (best ...


30

A pendulum is a day to day example of this. If you watch a pendulum swinging from left to right as it passes the mid point the velocity and acceleration are: The acceleration always point towards the mid point, so as the pendulum passes through the mid point the acceleration reverses direction but the velocity does not.


20

If $D^n$ denotes the $n$th derivative and $D^{-n}$ the $n$th integral, then we have that, $$D^n f(t) = D^m[D^{-(m-n)}f(t)]$$ providing $m \geq \lceil{n}\rceil$. For our half derivative, we choose $n=1/2$, and $m=2$, in which case we have, $$D^{1/2}f(t) = D^2[D^{-(3/2)}f(t)]$$ There is a general formula for the $n$th integral of a function, one of my ...


15

At least one mobile phone I've heard about uses an unbalanced spinning weight. As the weight moves in one direction, the phone moves in the other, in accordance with Newton's Third Law of Motion.


14

I am no mathematician and am a little afraid that my answer is too simple to be true, but here goes: I use Fourier transforms to define the fractional derivative. $x(\omega)$ is defined such that $$ x(t) = \int_{-\infty}^\infty \, \frac{\text{d}\omega}{2\pi} \text{e}^{i \omega t} \, x(\omega) \, .$$ Then any integer derivatives is $$ ...


11

Get on a car, turn it on and press the accelerator! (${{a}}>0$)... then press the brake (${{a}}<0$). Until the car is steady, the situation is as you described.


10

It appears to me the issue is understanding momentum conservation. An even cruder example would be to shine a bright torch out the back of your vehicle. Even though the photons have no mass, wouldn't the vehicle move forward? You also refer to mass in this manner in the paraphrasing of Newton's third law "proportional opposite mass/acceleration ratio ...


10

Here is a brief historical ideosyncratic intro to calculus. Calculus of finite differences Consider this problem from a typical IQ test: 2 5 10 17 26 ? What's the next number you expect in the sequence (this is not hard, you should do it). The n-th term in the sequence is given by: $$ n^2 + 1 $$ as you can see by substituting n=1,2,3,4,5, so the next ...


9

Yes. Actually photons exert pressure on any surfaces exposed to them. For example, photons emitted by the Sun exert pressure of $9.08 \mu N/m^2$ on the Earth. Reference: https://en.wikipedia.org/wiki/Radiation_pressure


8

These days planes measure their speed (and position) using GPS. In the old days (my father used to fly Tiger Moth's!) they would measure air speed for a rough guide, but correct their speed by spotting landmarks on the ground. In poor visibility it was not uncommon for pilots to get lost, sometimes resulting in tragedy when they flew into mountains or ...


8

Consider the frame of reference in which the rain is stationary; in this frame, you are moving upward (assuming no wind) at the rain's terminal velocity along with any horizontal motion you make. In this frame, the raindrop at a given point wets you iff you, at some point along your path, occupied that point; the amount of rain that hits you is proportional ...


8

Here is a very basic estimation: The kinetic energy of a 1000 kg car moving at 60 km/h is $$E=\frac{mv^{2}}{2}=\frac{1000kg(16.7m/s)^{2}}{2}=138.9 kJ$$ The heat of gasoline combustion is 47 MJ/kg = 35000 kJ/litre. Assuming 10% efficiency of the car's engine, you would need to burn $$\frac{138.9 kJ}{0.1\cdot35000kJ/l}=0.04 litre$$ of gasoline to accelerate ...


8

There's a simple way to look at this that doesn't involve any maths. Suppose the two cars are parked and are stationary, and you accelerate past them in your car. If you are accelerating forwards then from your perspective it looks as if the two cars are accelerating backwards (at the same rate). But the cars are at rest, so the distance between them can't ...


7

If you solve for $t$ in Eq. (5.1), and plug that into equation (1.1), you'll see that the solution looks like $x_B \propto v_A^2 sin(\theta) cos(\theta)$. The function on the right is symmetric about $\pi/4$, thus, as long as $\theta$ doesn't equal $\pi/4$, there will be two solutions (symmetrically about $\pi/4$). Of course, in general, there could be ...


7

This is all a complicated (and confusing, or just plain confused) way to say that, if you want the locomotive to pull the train, you don't want its wheels to slip. It's friction that prevents the wheels from slipping. I suggest you simply delete this sentence: This static frictional force, of the rails pushing forward on the wheels, is the only force ...


6

If you stand still you get infinitely wet. I think we can safely say that running is better than standing still. If you run at nearly the speed of light, then each drop of rain is effectively frozen in time, and you "carve out" an amount of rain equal to your body's cross-sectional area times the distance you travelled times the density of rain. If you walk ...


6

The distance from London to Australia is about 17,000km. If you wanted to minimise the acceleration you'd feel during the trip you'd accelerate continuously for the first half of the journey (8,500km) then decelerate at the same rate for the second half. To work out what acceleration is required you use the SUVAT equation: $$ s = ut + \frac{1}{2}at^2 $$ ...


6

Yes - you can even propel spaceships with it - Solar Sail Although Solar radiation pressure at the Earth is around 9E-6N/m2 while the thrust from a Saturn V rocket is 34 MN, so you would need a solar sail something like 2000Km on a side to get the same acceleration


6

My understanding of the question is that it's about minimizing the rate at which rain hits the car. That makes it different from this question, which assumes you want to minimize the total amount of water that hits you before you get to a certain destination. First let's assume the rain is perpendicular to the road and the car is a sphere. Then by the ...


6

Of course. Acceleration is the rate of change of the velocity. For motion in a line, if the object is slowing down, the acceleration is opposite the velocity. If the object is speeding up, the acceleration is in the direction of the velocity. Imagine you're pedalling a bike, gaining more and more speed and then, suddenly, stop pedalling and apply the ...


5

Absolutely. Acceleration is the change in velocity, so when you say that the acceleration reverses in direction, it means that the object is transferring from either speeding up to slowing down, like a skateboard which just passed the bottom of a U-shaped ramp, or from slowing down to speeding up. Think of the skateboard as having just crested a hill. ...


5

Newton's laws are testable. An example of a modern, high-precision solar system test of the first law is Battat 2007. The Eot-Wash group has done various high-precision laboratory tests, such as searches for coupling of spin to a preferred frame. The third law has been tested to high precision in Kreuzer 1968 and Bartlett 1986. For modern tests of the first ...


5

The drag on a moving object is given by the approximate expression: $$ F = \frac{1}{2} \rho v^2 C_d A $$ For my Ford Focus $C_d$ = 0.32 and $A$ = 2.12 m$^2$. The density of air at STP is about 1.2 kg/m$^3$, so at the UK motorway cruising speed of 70 mph (31.3 m/sec) the drag is: $$ F = \frac{1}{2} \times 1.2 \times 31.3^2 \times 0.32 \times 2.12 = 399 N$$ ...


5

If the total force $F$ on a mass $m$ follows Hooke's law, $$F~=~-kx,$$ then one can use Newton's 2nd law $$F=ma,$$ to infer that the motion is a simple harmonic motion $$ a =-\omega^2x, \qquad\qquad \frac{2\pi}{T}~=~\omega~=~ \sqrt{\frac{k}{m}}~,$$ cf. OP's correct belief. Now it only remains to solve the ODE $$ ...


5

If the person is moving the block in such a way so that the sum of the forces acting on it is equal to zero, how can he be moving it at all? Consider a person pushing the block of wood along a surface with friction where the force due to friction (a force proportional to the speed of the block) exactly cancels the pushing force from the person. The ...


5

The confusion arises because there are two different versions of what Earth's surface looks like, and two different models of how gravity works between the case where the ball goes into orbit and the case where both balls hit the ground at the same time. We often approximate gravity near the Earth's surface by saying that it is constant everywhere. This ...


5

You are correct: everything is in motion (or not) based on the reference frame. Motion is a relative concept, so you are never "moving" but only "moving with respect to something". Find a good basic primer here: http://en.wikipedia.org/wiki/Principle_of_relativity


5

Yes, with gravity and a generous definition of "moving".. it would be the same principle as the toys where you can control a sphere using a radio control (or using your iphone). The fish swims along the edge and gravity pulls it back down, which starts a rotation of the water and by friction to the sphere starts the rolling motion of the sphere on the ground ...


5

It is true that the double pendulum exhibits integrable behavior, when the initial angles are very small, however, in general, it is very difficult to characterize the chaotic behavior of the double pendulum in terms of the initial angles. There are other representations which provide a clearer picture of its chaotic behavior. The introductory section of ...


5

From a physicist's point of view the quote is nonsense: gravity and electrodynamics are how matter moves matter on the macroscopic scale and we have perfectly good theories for both of them. If the writer means "we don't actually know how it really works when you get right down to it; I mean not really know." then he's speaking pure, unadulterated ...



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