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Let us for simplicity consider the idealized case of a point particle with mass $m$, and let the time of collision be $t=0$. We choose upward (downward) as a positive (negative) direction, respectively. Let the velocity just before and after the elastic collision be $$-\lim_{t\to 0^{-}} v(t)~=~\lim_{t\to 0^{+}} v(t)~=~v_0~>~0.$$ Then the velocity, ...

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To reduce risk of slipping by increasing friction between tyres and road. while going up a mountain the opposing frictional force $F = μN =mg\cos\theta$ where $\theta$ is the angle of slope with horizontal .To avoid skidding $F$ should be large,so $\cos\theta$ should be large and hence $\theta$ should be small.so roads are made winding upwards.the road ...

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The weight of the ball cancels out the constraint force exerted by the Floor. But the ball does have some momentum due to its velocity which makes the ball bounce again.

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It appears that you are confusing the force you contribute (F1), with the force necessary to overcome friction (F2). The force necessary to overcome friction, is fixed by the mass of the object and the surface friction. As an example let F2 be 100N, lets assume you can only provide 80N, then you will not be able to move the object. If you get a friend and ...

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You're just comparing forces to forces, Newtons to Newtons. I guess if you want to be really nitpicky, you're talking about force per charge vs force per mass. But when one says that the electric force is much stronger than the gravitational force, one means that in dealing with the usual objects, like electron, the effects of gravity on the electron's ...

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The question is phrased in terms of dynamical concepts like force and mass, but there's a more fundamental kinematical answer that trumps these issues. If an object is moving with speed $u$, and you then apply a boost $v$, the object's new speed is not $u+v$ but rather $(u+v)/(1+uv/c^2)$. This is always less than $c$. Therefore it's not possible to ...

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In special relativity, you are better off thinking of things as four-vectors, rather than three-vectors. In that case, you generalize momentum to 4-momentum ${}^{1}$(and you take time derivatives with respect to the clock of the spaceship). Then, you have $\bf F = \frac{d{\bf p}}{dt}$ Since momentum is given by $p = \frac{mv}{\sqrt{1-v^{2}/c^{2}}}$, and ...

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The first thing I should point out is W$\cdot$x. x is the displacement. Direction is important. That's really not the most important thing for your answer though. The punch line, Work is done if there's a force (hence change in velocity) and it causes displacement So, if something's moving a constant velocity, it doesn't have any force acting on it ...

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A sailboat is moving at a constant velocity. Is work being done by a net external force acting on the boat? This is a bit of a trick question it tricks you into thinking there is a net external force. The boat is moving at a constant velocity; that's a given. That means that the net external force on the boat must be zero. But if I use Work = Force ...

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Good one. The net forward thrust afforded by power of the engine, wind, rowing etc., is constantly overcoming and exceeding the retarding force of resistance by water, air, etc. So, the movement of boat in water and air is constantly impeded or constrained by water and air resistance-tending to bring the boat to halt. The forward thrust is constantly ...

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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 ...

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As the Eiffel tower famously shows, a "good" self-supporting structure does not have a uniform section - instead, at every level the size of the supporting surface is large enough to support the weight of the structure above it without reaching a fracture / yield point of the building material. The melting point of ice is a function of pressure - so the ...

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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 ...

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To keep things simple lets say the tug of war rope is stationary, both sides are pulling with equal force. A and B are both people, A pulls left with 100N, B pulls right with 100N. Now in your second example we replace B with C, your immovable weight. A pulls left with 100N. Newtons law says that the block must be pulling right with 100N, otherwise the ...

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If you want to take the sum of moments, you have to take **all the external forces ** into account. The summation of moments has no preference for a certain force, allowing it to be in the equation. All forces are in the summation, only their resulting moment can be zero. If we look at the following example, assuming that all the spacings, e.g. ...

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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 ...

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The problem is that 9.81 m/s^2 is the acceleration during the fall, not the impact. Let's suppose the impact lasts 0.1 s when a ball hits the ground at 10 m/s, assuming inelastic collision (meaning, the ball doesn't bounce back), average acceleration DURING impact will be 10/0.1 = 100 m/s^2 rather than 9.81 m/s^2 and that's the acceleration you should be ...

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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 ...

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We would have to know more about the ice we want to build the wall with. For example, for ice in icesheets, you have an ice which effectively reaches a plastic region of the stress strain curve at around $0,5 MPa$. I am not a geologist, but I believe that the glaciers can be only thicker than $\sim 50 m$ thanks to it's specific shape and the fact that the ...

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It might not be best to say you'd feel it, as the oscillations of the sun's gravity (as due to rotations and revolutions) are relatively subtle here on earth. The moon's gravity has twice as much influence in terms of tidal force, which is probably the most prominent consequence of gravitational "stacking" in the macroscopic sense you had in mind.* Here's an ...

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Well given Newtons famous equation the force of gravity you feel is equal to $$F_{\textrm{gravity}}=G\frac{mm'}{r^2}.$$ It increases with the mass of the two objects being measured and is inversely proportional to the distance between the two objects in question. Strangely enough however, this technically stretches out nearly infinitely so the ...

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Technically, yes, however in the case of you standing on a planet: You feel the gravity of the planet and star, so you accelerate accordingly. However the planet also feels the gravity of the star, so accelerates towards it. So you only notice your acceleration towards the planet.

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You can analyze this by imagining a tiny tiny gap between the two masses. Physically we have exactly that, as the electrons at the surface of the first block are certainly not in contact with the electrons at the surface of the second block. Then we have a series of two collisions: the first is the initial impulse, the second is the collision between the two ...

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I'm going to just use $32^∘$ below; it doesn't make a difference. Your equation isn't correct. You should have $F_x−f_k = 0$ or $Fcos32∘−f_k = 0$. The x-component of F is $F_x = F.cos32^∘$. Writing $F_x.cos32^∘$ doesn't make logical sense. Why? Shouldn't I be able to use $∑F=0$ in this problem to find the answer? Yes, you can.

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You must obtain the magnetic field due to the solenoid according to the approximate calculations according to its dimensions. If the magnet is small you can easily found the torque, and also the force when it is in a region of non uniformity.

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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 ...

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You're mixing up different things. Two blocks of iron press your head more than one block because the Earth's gravity pulls two blocks stronger than one. This is why two blocks can tear through the paper where one can not. The molecular bit comes into consideration if you ask why the top block doesn't pass through the bottom one. This is because the ...

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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 ...

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If your friend's energy+ yours= F1, then you would see your own energy expenditure halved, which we know cannot be the case. If your friend helps you push the object, then you are no longer applying the same force, or (lazy answer) the force is no longer localised and motivates the part of the object most subject to friction. So, once the object is first ...

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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 ...

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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 ...

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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 ...

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Equations of motion explained: Sum of forces on body equals mass times acceleration of center of mass. $$\sum \vec{F} = \frac{{\rm d} }{{\rm d}t}(m\vec{v}_{cm}) = m \vec{a}_{cm}$$ Sum of moments on center of mass equals mass moment of inertia times angular acceleration of body, plus velocity related terms. $$\sum \vec{M}_{cm} = \frac{{\rm d} }{{\rm ... 3 Many students confuse the term work in physics with the conventional term of work. Your body wastes energy when you push something, and when that something doesn't move... 100% is wasted in the biological efficiency. 1st step: forget the concept of how hard it would be for you to do it. How much work is a table doing by holding up a 1kg weight? zero. It ... 3 The same force applied anywhere on a rigid object causes the same translational accelleration. The difference is that forces not applied in the direction of the center of mass will also cause some rotational accelleration. Remember that F and A in F=mA are vectors. You can therefore treat the vectors as components in any orthagonal system you like. One ... 0 It is easy if you draw the Free body diagram of the box. The FBD should show only the EXTERNAL forces. There are 2 external forces. First, the force of the earth's gravity ON the box which equals Mg and acts in a downward direction (here g = +ve). Second the force of the table ON the box which is N acting upwards. Now you decide on your sign ... 3 In a sense your question is entirely apropos. The critical distinction to be made, however, is that the two forces can only be summed to zero if we are talking about the two-body system. Each body, considered in isolation, experiences an unbalanced force and thus experiences accelerated motion. The system, however, only experiences mutually-canceling ... 1 This may be cheating, but I think the problem is easier if you use conservation of energy. If you set the gravitational potential energy reference to the height of m1, then initially you have$$ 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} $$A bit of ... 1 I worked the problem out a ways and it involves quite a bit of tedious algebra. The technique I used was to include the electric force in the Fx and Fy equations by looking at the angle that  \hat{r} (the vector between the two charges) makes between the charges. For example, the equation i came up for Fx is T_x - \frac{Kq^2}{r^2} \cos{\theta} =0 where ... 0 This is the classic horse and buggy brain teaser. If the horse pulls on the buggy, the buggy must pull back on the horse with an equal force so who moves anywhere? The answer lies in the fact that the buggy pulls back on the horse because it is being ACCELERATED. That is where the force comes from. In the case of the two boxes, the second box (one at the ... 1 A better electrical analogy to Newton's second law might be inductance:$$ 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 ...

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I'm not sure if I'm just misinterpreting your question, but we can most certainly influence the acceleration of an object directly. To do this we must take into consideration what is actually happening as we accelerate; Acceleration, in physics, is the rate at which the velocity of an object changes over time. While it is a sum of the net forces over the ...

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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 ...

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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 ...

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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 ...

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If the acceleration (presumably measured on the sled) is the same, then the mass of the sled has no effect. It will take 40 times a s much force to accelerate the 40 lb sled. What is more likely to be happening is that at a particularly frequency, then mass of the cap and the spring constant of the legs will be such that you get metal fatigue from the ...

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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 ...

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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. ...

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A known mass tied to a thread will exert a constant force, proportional to its mass. That's the easiest of all. A spring under a light load will create a constant force that is proportional to the extension. See Hooke's law. A magnet and a piece of magnetic iron (or other magnet) tied to a known distance will do, too. If it is an electro-magnet, the force ...

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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 ...

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