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1

As dmckee's comment says, your question doesn't really have an answer. However it's exactly the sort of question that fascinated me in my time as a teenage physics enthusiast, and I think it touches on some really interesting aspects of physics. To get a grip on this you need to understand how matter is described by quantum field theory. This is not ...


-2

I am not interest to this topic please change the current topic.


0

I am going to assume that you are well aware of and and are comfortable with rotational mechanics of rigid bodies(if you aren't, then you must give the chapter a read at once). I will briefly try to explain it. Consider three points on the ball, the point nearest to the center of the ramp, the center of mass of ball and the point farthest to the center of ...


0

I don't think anyone really knows what matter is.It's abstract.We just know its properties.


1

Braking acts to stop the front tire. Friction acts at the contact patch under the front wheel to introduce a vector force directed towards the back of the bike. Since the force is not directed through the center of mass of the motorcycle/rider system, it introduces a moment or torque that acts to rotate the motorcycle and rider such that the back tire begins ...


0

In order to explain the forward motion of the horse-cart let us first consider the forces acting on the horse & the cart. Two forces act on the cart: i) Tension $T$ on the rope connected to it. ii) Friction $F$ on its wheels. Again, the two forces acting on the horse are: i) Ground reaction $R$ whose horizontal component pushes the horse forward. ...


0

The system horse+carriage is in equilibrium. However the horse's feet (burning calories) will generate a force on the floor. By third law the floor will generate a force on the system horse+carriage. The most interesting thing is that the floor will move backwards as well! This is. If the horse+carriage are on the Earth, when it moves due to calories being ...


3

The braking force acts between the tyre and the road. The centre of mass is above this point so there is a rotational effect which increases the force going down through the front tyre and decreases the force going down through the rear tyre. Because the amount of braking force the tyre is able to produce is limited by the amount of force going down through ...


0

For the top surface of an object to be flat and level, the object below the surface must be under compression. It would be possible to have a walker's tightrope that was almost perfectly flat and level on the top if there two additional ropes rope below and to either side of the first one, and the lower ropes were separated from the first one by spacers, ...


0

In ideal conditions, if the geometrical objects are such that when they collide the collision doesn't generate rotation of the objects, then Newton's Cradle shouldn't be different. The bodies will exchange momentum just as the spheres do. Notice that contact area doesn't matter if you think of the collision as an interaction between point-like particles. If ...


2

Lots of excellent answers here, but for fun, lets think about this backwards. Imagine you have the worlds first and only FRONT wheel drive motorcycle, and your rev it up and pop the clutch. What kind of launch do you think you would get with very little weight on the front tire? The reverse is true during braking when the deceleration shifts the weight of ...


0

Recently, I have been thinking about alternative causes for the rotation of our planets. You're sixteen. I've noticed that while people of your age can understand Newton's first law of motion, they don't understand the rotational analog of this law. Just as an external force is needed to change an object's momentum, an external torque is needed to ...


0

A CCD is a charge couple device. You can think of it as like a digital camera screen. The detector is therefore made up of pixels of size $\Delta y$. You must find the minimum difference between $v_1$ and $v_2$ such that the two particles hit the detector with a separation at least $\Delta y$. In part b) of the question you should have worked out $y$ as a ...


0

A CCD is a charge coupled detector - a photosensitive (usually) pixelated device. The "charge coupled" has to do with the way signals are transmitted to the readout electronics. For the purpose of this question you only need to know that this detector can resolve particles when they are $\Delta y$ apart - because those particles will hit different pixels on ...


2

For the record, here's a worked solution: If $r$ is the distance between the two point masses $m_1$ and $m_2$--which start at rest--then as both accelerate towards each other, where $$ \frac{d^2r}{dt^2} = -\frac{Gm}{r^2} \ \ \ , \ \ \ \ \ m = m_1 + m_2$$ The first step in solving this equation is the least obvious: Multiply both sides by $\displaystyle ...


0

I'm not going to give you the answer, but try eliminating the velocities by substituting information provided to you about the momentum relationship. The velocities should not be in the final solution.


0

I think the question is trying to guide you to the "conventional" formulation of the damped harmonic oscillator: $$\ddot x + 2\zeta \omega_0 \dot x + \omega_0^2 x = 0$$ Where the factor $\zeta$ corresponds roughly to your $\epsilon$. See http://en.wikipedia.org/wiki/Harmonic_oscillator Does that help?


1

Calculate the kinetic energy of the water coming out in unit time. That is the power you need. $$Power = \frac12 (\rho A v) v^2 = 31.4 kW$$


0

Suppose one has a piece of thin string which is sitting on the ground in front of the wheel, wraps around the wheel once, and then continues along the ground behind the wheel. As the wheel rolls, different parts of the string will be wrapped around it. At any point where the string is in contact with the wheel, its velocity will match that of the wheel. ...


1

Any change in speed is acceleration. Accelerating is equivalent to gravity being different. Braking slows you down, which means you are accelerating backwards, which means you feel gravity pulling you in the same direction as on a downhill slope. As on that downhill slope, more of the your bike's weight is supported by front wheel and less by the back wheel. ...


6

It is great that you "think differently" about problems - that is at the heart of all innovation. When it comes to the rotation of planets, you have to go back to the origins of the solar system: Planets are formed by accretion: a large cloud of debris starts to experience some gravitational pull, and as one "lump" becomes bigger than the others, it starts ...


0

Mathematically you can of cource use either form or $\epsilon$, but the choice may depend on the interpretation of $\epsilon$. Personally from the formulation "when c is small $\epsilon$ is small", I would go with the linear one, since the quadratic would be: "when c is small $\epsilon$ is really small".


1

In a system, the total sum of forces when added together equals mass times acceleration: $$ \sum F = \frac{\mathrm{d}p}{\mathrm{d}t} = \frac{\mathrm{d}mv}{\mathrm{d}t} = m\frac{\mathrm{d}v}{\mathrm{d}t} = ma $$ Since the sum of the forces on the robots is zero, there is no acceleration. However, the tension of the string is not contingent on the movement. I ...


32

Using the brakes on the front of the bike causes your weight to shift forward. Additional weight allows more force before the tire will slip (skid). If you brake hard enough the back tire of your bike will lift up and at that point all of the mass is distributed on the front tire. Remember the maximum force is $F_{max} = \mu F_{normal}$ and $F_{normal}$ ...


0

Notice that water and trolley are part of a system. If water is flowing out at a constant rate (or not) its momentum will be changed only by an external force, say friction with the floor when the water reaches it. If you consider the trolley as just the trolley, its momentum and kinetic energy won't change. If you consider the system as a whole you need to ...


1

If the force applied is greater than the friction, it just means that the object will accelerate. Some of the work goes into overcoming friction, the rest goes into accelerating the object (and thus kinetic energy). Work done by the force (15 N in your case) is just force times distance - it doesn't matter how that work is then split between the friction and ...


0

Take the reference frame as centered in the fixed axis. The $R$ that connects the origin to the centre of the spinning disk forms an angle $\phi$ with the horizontal. Now, inside the disk of radius $r$, the angle of a certain point mass is given by the angle it forms inside the spinning circle, which we'll call $\theta$. Now take as generalised coordinates ...


1

If something is in freefall, starting at $v = 0$ at height $h$, then $$\begin{align} a &= - \frac{GM}{r^{2}}\\ v\,{\dot v} &= - \frac{GM v}{r^2}\\ v\,{\dot v} &= - \frac{GM {\dot r}}{r^2}\\ \frac{1}{2}v^{2} &= \frac{GM}{r} - \frac{GM}{h}\\ v &= \sqrt{2GM\left(\frac{1}{r} - \frac{1}{h}\right)} \end{align}$$ Then, $$\begin{align} J ...


0

The answer depends on the ascent rate $\dot h$. Differentiating gravitational acceleration with respect to time yields $$\frac {d\,g(h(t))}{dt} = \frac {d}{dt}\left(\frac {\mu_E}{(R_E+h(t))^2}\right) = -\frac {2\mu_E}{(R_E+h)^3}\dot h = -\frac{2g(h)}{R_E+h} \dot h$$ where $\mu_E$ is the Earth's standard gravitational parameter, $\mu_E = GM_E$. It's better to ...


0

While it may be interesting and instructive, you don't really need to know much about the motion of the jumper. You can answer the quesetion by using a simple energy balance. At the bottom of the fall where the velocity of the jumper is 0, all the potential energy of the jumper from there to the starting point has been absorbed by the spring: potential ...


2

From the moment that the spring (bungee) comes under tension, the motion of the jumper can be described as a combination of simple harmonic oscillator and a constant acceleration - that is, a sine wave with an offset (the offset is the "equilibrium point" you mentioned). But you don't need to go there. If you say that "at the bottom of the drop" the entire ...


0

There is not enough information here. We need to know how far it is being stretched. However, I will show you the formulas needed. $$ F = \frac{1}{2}kx^2 $$ Where $k = F/d$ so: $$ \frac{1}{2}\left(\frac{F}{d}\right)x^2 $$ $x = $ new distance (in meters) $F = $ force (800) $d = $ distance (0.2)


1

Quick answer, the ground. Try to use a lever on unstable ground (like mud) see what happens.


0

Historically, the use of levers was well understood long before Newtonian mechanics - the scales that merchants used for commerce were all based on levers, and any time accountants are involved, the science will be very well understood. As such, Newtonian Mechanics had to conform to the relationship between the ratio of weights and lengths, not than the ...


0

Do you know about conservation of energy? Do you know about conservation of momentum? If you do then here's what you do:: Conserve energy for the entire system (Including velocity of the wedge). At the lower point the potential energy of the smaller block changes (potential energy= mgh) here is one relation between the velocity of the block and the wedge. ...


1

2 is correct and so is 1. force is invariant under change in frame. However work varies under change of frame.This is because the displacement of a body changes with change in frame. For e.g., a body moves 5 meters with respect to a stationary body. However if there exists a frame that moves 2 meters at the same time then according to that frame the previous ...


0

Your first question is answered quite easily - the force may be the same in both frames, but the distance traveled is not. In the frame where the change in KE is greater, the object will also cover a greater distance. So the change in KE is not frame invariant, and neither is the work done.


0

What does the principle of the lever state? it states that there is a compatibility condition $\frac{F_2}{F_1}=\frac{r_1}{r_2}$, related to torques (which, transposing, $F_2 \cdot r_2 = F_1 \cdot r_1$ is another form of the conservation of energy). See a related question for the relation between torque and force. So the lever compatibility condition (for ...


2

Why didn't Newton just propose the 2nd Law (F=p˙) and leave it at that? The 2nd Law implicitly contains the first, doesn't it? If so, it seems he wasn't following his own Rule #1 of Book 3 of his Principia: "We are to admit no more causes natural things than such as are both true and sufficient to explain their appearances." Simply because ...


2

I assume that by $M_3$ you mean $I_{zz}$ of a diagonal inertia tensor. Nothing wrong with your notation though, I'm just clearing that up. The inertia tensor plays an analogous role to that of mass on linear motion. If we are talking specifically about $I_z$, it means that IF the body has angular velocity around this axis z, the kinetic energy associated ...


2

I think it refers to the ability of a right handed player/ left handed player to move more in one direction than in other. Nothing to do with physics. Even if it is meant to be scientific its totally wrong .Germans aren't always correct.


0

I'm not sure what he meant by that but he could be referring to an aerodynamic effect. The rotation influences how air flows around it, creating forces, just like lift in a plane. See, for example, Magnus effect: http://en.wikipedia.org/wiki/Magnus_effect Of course, the direction of the curve isn't necessarily 'left' but it depends on the rotation.


2

At any given moment, the pendulum is swinging in a certain plane. The driving force should be within this plane. If the earth weren't rotating, then such a force could never cause the plane of swing to rotate about a vertical axis, since by symmetry there would be no preferred direction for the rotation.


1

Since gravity is constantly acting on both spheres once released, both objects have constant downward directed acceleration and, thus, have zero velocity only for infinitesimal time. That is to say, the velocity of both objects is not constant at any time between their release and their impact with the ground. Assuming the objects have different upward ...


0

You seem to be having quite the fun with this particular diagram over the past few days. The tension in the string is only equal to the weight of the hanging mass if the system is in equilibrium. Other answers have said that if the system is at rest, $T$ and $m_{hanging}g$ are equal, but that's only partially true. The key is that the system has to be in ...


-1

The mass of the body on the table is not directly important - the important thing for the tension in the string is the force of friction $F_f$ in the diagram. So it depends on how smooth the surface of the table is. If $F_f$ is greater than or equal to $mg$ then the tension $T$ is $mg$ and the system is at rest. If $F_f$ is less than $mg$ then the ...


0

It does not depend on the mass but on the friction of the horizontal surface. If friction is high enough so that the masses do not move, the answer is yes. But if the friction is zero the masses will move (regardless of how large the mass on the table is) and the tension will be less than the hanging mass.


0

You can use angular momentum conservation because friction is an internal force. Angular momentum of a system changes if there is a net external torque on that system.


3

The $L\cdot\Omega$ term comes directly from the change of frame of reference, especially from the transformation from the static frame of reference to the rotating frame of reference. Let $\mathcal{R}\equiv(x,y,z)$ the initial static frame, and $\widetilde{\mathcal{R}}\equiv(x',y',z')$ the rotating frame at a constant velocity ...


2

If the "foot" of the incline is itself also inclined, you need to take into account further increase in energy due to gravity. If the foot is horizontal, then your approach is fine - because you compute the normal force times coefficient of friction to get force of friction, and force times displacement is work done by the object. When it runs out of kinetic ...



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