# Tag Info

32

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 ...

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

In 1619, almost a century before Newton published his groundbreaking Principia Mathematica, Johannes Kepler made a revolutionary contribution to observational astronomy. He noticed that the square of the orbital period $(P^2)$ (the time that it takes for a planet to go around the sun) of a planet's orbit is directly proportional to the cube of the length of ...

8

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 ...

7

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 ...

7

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 ...

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 ...

6

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

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 ...

6

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 ...

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: $$s = ut + \frac{1}{2}at^2$$ ...

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

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

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 ...

5

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 ...

4

The relationship is between speed, distance, and the angle thrown. The distance the ball travels before coming back to the same height is further if you throw it faster, but less if you throw it at a lower angle (up to about 45 degrees). By adjusting both the speed and angle of your throw, you can have two throws that go the same distance at different ...

4

The bullets don't hit the ground at the same time exactly because it is very difficult to fire horizontally, there is air resistance to account for, the ground may be sloped, there are Coriolis forces, etc. Usually, when people refer to this phenomenon, they're referencing the principle of Galilean relativity. You can read the famous excerpt from Galileo's ...

4

No. You, your bottle and drink within it, all (regardless of the bottle orientation) move with the same acceleration that amounts to gravitational acceleration. In case you are standing on the ground, you and bottle are at rest, while drink (neglecting the constrains of the bottle) moves with gravitational acceleration toward your thirsty throat. :) Edit: ...

4

"I seem to think that impluse is high if for same force, time is higher." This is correct, to develop an intuitive understanding, we must just realize that impulse simply refers to a change in momentum. So lets say you apply a force of 1N to an object. Is the object going to have a larger change in momentum if you apply 1N for 1s or 1N for 5s? It ...

4

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: $$F = \frac{1}{2} \times 1.2 \times 31.3^2 \times 0.32 \times 2.12 = 399 N$$ ...

3

Mark came the closest in his answer, but I want to address some of the deviations from idealism in greater detail. People have mentioned the angle the gun is at and air resistance, so I'll try to touch on both. You have a height $h$, say the velocity of the bullet is $\vec{V_b} = < V_{bx}, V_{by}>$ (add $t$ dependence when necessary). The time it ...

3

This is indeed true. The easiest way to think about it is in planes. You have horizontal and vertical planes for velocity and acceleration. For both bullets, there is initially no vertical velocity and the only net force acting upon them is gravity. You would then expect them to act similarly in the vertical plane, which is why they both hit the ground ...

3

There is a question of what you are trying to learn about friction from this convoluted example. What may be slightly confusing is that there are two types of friction static friction and kinetic friction. Kinetic friction is the friction associated with two substances sliding against one another, which can only happen when one is moving relative to another. ...

3

Well, I am pretty sure Kepler is not docking with ISS, because Kepler is orbiting Sun and ISS is in low Earth orbit. But in general, docking is performed over several days with slow adjustment of velocities so orbits match with final stage docking performed with small thrusters usually called linear RCS. What is important when docking is relative velocities ...

3

The answer is that you are using a Discworld assumption (i.e. Earth is a plane) to derive conservation of momentum in the horizontal directions. Motion in this direction is then decoupled from the motion in vertical direction and so both balls have the same altitude at all times. As you might imagine, Earth is round, so the above assumption fails at large ...

3

If you are falling you feel weightless even though you are still being affected by gravity - the sensation of weightlessness is just that there is nothing around you to press against you. An astronaut in orbit is effected by gravity almost as much as you are (being 10% further from the centre of the earth doesn't make gravity much less)

3

Every motor has a torque vs. speed curve, and the product of those is power, and it has a speed at which it generates maximum power. Every propeller generates thrust roughly proportional to its speed squared, or its power cubed. So for a given power, it has a particular speed at which it uses that much power. What you want to do in an airplane is find the ...

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