Tag Info

Hot answers tagged

35

If you're pushing a 10-ton truck and it's not moving, you are not doing any work on the truck because the distance $ds=0$ and the nonzero force $F$ isn't enough for the product $F\cdot ds$ to be nonzero. Your muscles may get tired so you feel that you're "doing something" and "spending energy" but it's not the work done on truck. You're just burning the ...


15

At the risk of oversimplifying: bonds form because the bonding electrons can lower their energy by being attracted to both nuclei. Take the simplest possible example of H$_2$. The electron density looks like: (image from Hyperphysics). The formation of the H-H bond increases the electron density between the two protons, and the electrons in this region ...


5

For the case that you have drawn, the behavior of the drop is actually the exact opposite of what you mention: it will move from right to left. This is caused by surface tension and the curvature of the droplet caps which creates a larger pressure in the drop at side B than at side A. To make it more quantitative. Let's assume that the funnel is ...


5

Your observations are spot on. I usually write Newton's second law this way: $\vec{a} = \vec{F}/m$. This form makes it clear that the law is a relationship between the dynamic variables force and mass, and the kinematic variable, acceleration. $F$ and $m$ describe the situation, $a$ is the result. Cause and effect, if you will. In fact, that's one ...


4

The wind is certainly doing work, because it applies a force and the point where the force is applied is displaced. However it isn't doing any work on the boat, it's doing the work on the water. The key point is that the net force on the boat is zero. We know the net force on the boat is zero because the boat is moving at constant velocity - if the net ...


3

Many students confuse the term work in physics with the conventional term of work. Your body wastes energy when you push something, and when that something doesn't move... 100% is wasted in the biological efficiency. 1st step: forget the concept of how hard it would be for you to do it. How much work is a table doing by holding up a 1kg weight? zero. It ...


3

The same force applied anywhere on a rigid object causes the same translational accelleration. The difference is that forces not applied in the direction of the center of mass will also cause some rotational accelleration. Remember that F and A in F=mA are vectors. You can therefore treat the vectors as components in any orthagonal system you like. One ...


3

In a sense your question is entirely apropos. The critical distinction to be made, however, is that the two forces can only be summed to zero if we are talking about the two-body system. Each body, considered in isolation, experiences an unbalanced force and thus experiences accelerated motion. The system, however, only experiences mutually-canceling ...


3

I'm sure everyone has had that concern when we encountered the definition for the first time, in school. There is a valid reason why this definition is still persisted with, despite the deficiency that you hit on. The most popular (and simple) forces in physics (also the ones with which we begin learning physics) are conservative forces, implying that the ...


3

You should read the article in wikipedia on nuclear force. Various models exist that describe the behavior of nuclear forces, which are the result of a spill over of the strong force, the force that exists within the proton and the neutron. From the link Force (in units of 10,000 N) between two nucleons that experience the nuclear force, as a ...


3

Because the kite thread is never vertial, it is inclined in the direction of the wind: So the wind will push your paper in the same direction, and the thread will force it to go upwards. For more detail, in the paper there will be 3 forces: Gravity. It pulls down. Wind. It pushes to the right. Reaction of the thread. It will force the paper to move in ...


3

"but how can you have a "basis" that changes at every point?" This is really the root of your problem. Mathematically (and physically) speaking, such a basis works fine. You just have a (hopefully temporary) conceptual problem. Maybe try thinking of it this way: How does $\hat{x}$ know to point "to the right" in cartesian coordinates, at an arbitrary ...


2

I think I see your question, if the surface of the air bubble were perfectly flat, and the air + cup didn't float, then the surface would have an equal pressure across it and it would not move. The system, however, is in an unstable equilibrium, the slightest perturbation will cause the bubble to rise out of the cup. Consider this, since the pressure of the ...


2

There is, effectively, only gravitation and friction acting on the pack of gum. However, the friction is not that strong (it is mostly independent of the velocity of the book, and dynamic friction is weaker than static friction) and it doesn't have that much time to act. Hence it doesn't affect the momentum of the gum noticeably. This is very related to the ...


2

Perhaps an easy way to see where Newton's second law comes from is as follows. Imagine an object sitting in space (so no friction, etc. to worry about). You can push on it (exert some force on it), and see what happens to it. When you push on it with a constant force, you see the object start to accelerate, which means its velocity increases linearly with ...


2

Let's work through this problem... $$\text{Power} = \frac{\text{Work}}{\text{Time}} \; \text{and} \; \text{Work} = \text{Force}\cdot \text{Displacement} $$ therefore $$\text{Power} = \frac{\text{Force}\cdot\text{Displacement}}{\text{Time}} $$ From Newton's Second law we know $$\text{Force} = \text{Mass}\cdot\text{Acceleration}$$ Substituting again: ...


2

You have a chain of action and reaction. There is twice the weight on our head because the forces felt by the lower block are its weight, plus the action of the top block (minus the reaction force of your head). Then how is it on a molecular scale? Well, just the same: if you imagine a crystalline solid with horizontal layers, one layer feels the action of ...


2

When you take a brass plate of considerable thickness and place it in between two charges, say positive and negative, induction takes place in the brass plate since it is a conductor: the electrons shift to the end near the positive charge while the cations stay near the negative charge. Now, induction occurs in order to make the field outside a certain ...


1

This may be cheating, but I think the problem is easier if you use conservation of energy. If you set the gravitational potential energy reference to the height of m1, then initially you have $$ U_i = k\frac{q^2}{d}. $$ The final energy will have a gravitational potential energy and an electrical potential energy: $$ U_f = mgh + k\frac{q^2}{r} $$ A bit of ...


1

I worked the problem out a ways and it involves quite a bit of tedious algebra. The technique I used was to include the electric force in the Fx and Fy equations by looking at the angle that $ \hat{r}$ (the vector between the two charges) makes between the charges. For example, the equation i came up for Fx is $T_x - \frac{Kq^2}{r^2} \cos{\theta} =0$ where ...


1

A better electrical analogy to Newton's second law might be inductance: $$ V = L \frac{\mathrm d i}{\mathrm d t} $$ The only reason this looks different is that physics has a name and conventional symbol for the derivative of speed, but electronics does not have a name for the derivative of current. So let's just pretend that the word acceleration does not ...


1

We should think a bit more carefully what force, mass and acceleration really are: For simplicity, we consider a classical point particle. The force $\vec F$ is something externally applied to the particle, it is a property of its environment. Most often, it is the gradient of a potential, $\vec F = - \nabla V$, but it need not be. In general, it is some ...


1

In my view, an expression like V=RI, practically speaking, isn't much different. You have to be concrete with a real-world example. I can connect a resistor across a voltage source, and then I can only alter the voltage or change the resistance value to change the current; I cannot change the current directly in this configuration. Your statement that ...


1

The brass plate is a conductor, so the potential will be the same on both sides. The thickness of the brass plate therefore subtracts from the effective distance between the two charges, making the electric field strength higher in the remaining open space between the charges. This stronger field will cause more force to be experienced by each charge. ...


1

There is no quick answer, except if the droplet is completely non-wetting or if it is at least partly wetting. If it is completely non-wetting, it will be move towards the wide side of the funnel until it is a spherical drop touching only its wall. If it is at least partly wetting, it will move to the narrow side until it reaches its apex (if air is ...


1

By way of analogy, think of what happens when you blow up a balloon and let it go. It spins around, goes this way and that. A balloon rarely goes straight, without spinning. The thrust from a balloon rarely goes through the center of mass. It rotates and translates. Because the thrust vector itself turns with the rotating balloon, the translation is not ...


1

Consider this container in pressurized air but zero-gravity (and ignore surface tension, which would make the liquid ball up). If your guess were right, the liquid would squirt out the small hole on the right, but that ignores the role of the wall on the right, which counters the pressure on the left. Think of the liquid as a collection of horizontal ...


1

When the cup is tilted up, the water wants to flow into the cup. That is what water does - it attempts to flow downstream. In doing so it displaces the air. Now the air experiences the force of the water (pressure below bubble > pressure above)


1

Equations for magnetic interactions between objects tend to be a lot more complicated than for electrostatic forces. This is because while electric fields are produced by and exert forces on charge, a scalar, magnetic fields interact with electric currents, the flow of charge in a particular direction, which is a vector quantity. This makes the equations for ...


1

Say you have a weight tied to each side a a rope which is strung over a pulley with friction. Here's a really easy way to see why the tensions on each side of the rope can't be equal. Imagine a really stiff pulley - in other words, ${\bf F}_{friction}$ is high. If that's the case, it'll be possible to balance unequal loads on this pulley system - i.e. a ...



Only top voted, non community-wiki answers of a minimum length are eligible