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How can we detect Earth's spin? Apparent motion of Sun You will have observed that the sun reappears every 24 hours. There are two common explanations for this. One of them is that the earth rotates with a period of approximately 24 hours - this is the only explanation supported by the scientific evidence. The main alternative had a rather convoluted ...

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Potential and potential energy are defined for pairs of objects, not individual objects. It's meaningless to say "the potential energy of A". One must say "the potential energy of the system consisting of A and B". There is only one potential and potential energy in your problem. Perhaps the confusion comes from the way potential is introduced in ...

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Here is the proof: Please refer to the wikipedia page on eccentric anomaly for a diagram and a couple of intermediate formulae. For an ellipse with the usual formula $x^2/a^2 + y^2/b^2=1$, it is the case that $\sin E = y/b$, and also by studying the figure on the wiki page you can see that $\sin (\pi-\nu) = \sin \nu = y/r$. Thus the two results you wish to ...

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It is independent of the mass of the planet if you assume the bearings are frictionless. Also assume that the tyres do not make dents in the ground. However, in reality, the bearings have friction. Additionally, there is rolling resistance as the tyre makes small deformations in the ground as it is rolling. This is why you see a characteristic "W" shape in ...

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If we assume all other things being equal other than the downward force due to gravity, the vehicle on Earth would be capable of greater acceleration. The ability of the tires to grip the surface on which they are resting depends on the downward force keeping them in contact (Coefficient of friction). That will pretty much relate the gravitational ...

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Always, always, always start problems like this by drawing a diagram: This make it obvious why cos and sin are used as they are.

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Mass is not a function of time, so the only thing inside the integral that needs differentiating is the position. Let me try to make this clearer. The integral over $m$ is really the integral over the volume: $$\int_m r \cdot dm = \int_V r \cdot\rho(r')~ dV$$ Here I deliberately distinguish between $r$, the vector from the origin of the coordinate ...

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You can't get an exact number for this without some assumptions. However, you can develop a relationship. What we do know is that one full period corresponds to some amount of time (my niave guess would be 1 second but I believe I have heard faster clocks which might be half seconds). So we introduce some constant C which is the amount of time reported on ...

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You need to remember that there are two forces acting on the pendulum: (1) Gravity toward the earth, and (2) Tension toward the center of the circle formed by the arc that the pendulum describes. If gravity is resolved into a vector perpendicular to the arc, and a vector tangential to the arc, the tangential component is a restoring force that returns the ...

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Yes the bullets can fall down and injure or kill you. In fact in countries were celebratory gun firing is possible people are often injured by falling projectiles. Shooting straight up is less dangerous than at an angle because the terminal velocity is much lower than the muzzle velocity of the projectile. When shooting at an angle some of the horizontal ...

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In a collision momentum is conserved if there are no external forces. enefgy is also cionserved but in the examples kinetic energy is an important parameter. Elastic collisions are ones when kinetic energy is conserved. In non-elastic or inelastic collisions kinetic energy is not conserved and some kinetic energy can be converted into heat, sound and in ...

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Momentum, energy, angular momentum, and charge are conserved locally, globally, and universally. One must remember that conservation locally (within a defined system) does not mean constancy. Constancy occurs only when the system is closed/isolated from the rest of the universe. Conservation means that these quantities cannot spontaneously change. Let's ...

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Perhaps the least convenient but the most direct is go to the Moon and observe the Earth. The (average) length of day as measured by timing stellar transit to stellar transit, sidereal day, differs by 4 seconds from the length of day defined by timing noon on one day to noon on the next day, solar day. A satellite launched East requires less energy to ...

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Yes, it is a sort of contradiction. More precisely, it says that you cannot model water's liquid-to-gas phase transition with an ideal gas model, and any attempt to do so will be fraught with contradictions. Now as to some misunderstandings that you may have: first, be careful about thinking of potential energy as infinite at infinite separation: that's ...

1

This is because $P(t)$ as stated is the instantaneous power as a function of time and $W =\mathbf F\Delta \mathbf x$ holds only for constant forces. More generally, recall that a definition of work is the integral: $$W = \int_C\mathbf F(x)\mathrm d\mathbf x$$ Where $C = C(x,t)$ is some curve in space/time. Expressing in terms of time gives: $$W = ... 1 For a force to do work, it must act upon an object as that object moves some distance (with a component along or opposed to the direction of that force). Holding an object stationary, but carrying the strength of a given force, adds no more work done. Because of this, we say that$$W=\int_C\vec{F}\cdot d\vec{r}$$where C is the curve in space along which ... 1 Think about the problem from the point where you neglect the wire and just analyse the motion. You have gravity and at any other point the force has to balance gravity that you end up in a circular motion. In A its towards the origin whereas in B there is no force (except gravity). Different from the classical space-orbit or car-in-a-curve problem is ... 1 You should use 1. Conservation of energy and 2. Centripetal force to the problem. I think you can figure out that when the angle subtended by the bead along with x-axis is between 0 and sin^{-1}(2/3), the normal force acting on the bead due to the wire is inwards. Hence, the normal force on the wire would be outwards. It is an easy problem, so I am not ... 1 The small object exerts a force in the opposite direction to the normal force on the cart 1 There is always air resistance, unless you are in space. If there was no air resistance, a bullet would land at the same speed it was shot at. On Earth, a spent bullet is rarely lethal. In big wars (like World War II) it was common for soldiers to get hit by spent bullets. Normally they will hit you and fall away, causing just a bruise, An unlucky hit ... 1 Since the springs are connected in series, they will experience an equal amount of force as the tension is same inside both the springs in equilibrium. If the tension is non-zero, then some part of spring will accelerate and therefore equilibrium will be destroyed. In case of parallel springs, the force exerted by them will be the sum of the two forces as ... 1 As we know, for a mechanical system being in equilibrium means having total force equal to zero. Let's look now at the second spring on a picture b): to be in equilibrium, force from the first spring F_{k1} (action of the first spring on the second) should equalise external force F_{out}. Thus, F_{out}=-F'_{k1}. This, in turn, after applying Newton's ... 1 The magnitude of centripetal acceleration is \frac{v^2}{r} instantaneously. It applies no matter the speed on your circular path. (Technically it's true for any curve, but r would be changing on non-circular curves, making calculations more difficult.) The tangential acceleration is constant, so you can write a function for v. Then you have two ... 1 As Nogueira stated, the air is made up of more than one particle. If the idea you're using is that a force occurs between two objects, then you'd have to treat each air molecule as an individual object, and you'd have many, many forces, each between the skydiver (who you could treat as a single body) and an individual air molecule. Of course, this gets messy ... 1 You're only a half-step away. You listed conservation of energy and linear momentum, both of which are due to there being no external forces on the three-charge system. But with no external forces, you know that the center of mass of the system won't accelerate. Since the COM starts at rest, this means that the COM will remain stationary. Think about what ... 1 Assume that each swing advances the second hand by one second on the dial. You have a pendulum which is supposed to have a period of exactly one second. If the pendulum has got longer in length it will have a longer period and so it will take longer for the pendulum to advance the second hand by one second on the dial. So the clock will run slow. Just ... 1 You are on the right track. For a mechanical pendulum, the relationship is linear. You don't need to know how many swings of the pendulum corresponds to how many seconds. If the pendulum is x% slower, it will report x% fewer seconds per day. Now since length goes as \ell = \ell_0(1+\alpha \Delta T) and period of pendulum as$$T = ...

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I have an old pendulum clock (probably over 100 years old) still in working order. I'm pretty sure that the hands advance linearly with the number of swings of the pendulum. I keep my house about 10 degrees Fahrenheit cooler in the winter than in the summer, but don't really notice a difference in its timekeeping. The bob is suspended by a wire made of ...

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If a ball of say radius $R$ rolls without slipping it has both linear ($p$) and rotational momentum ($L$): $$p=mv$$ $$L=I\omega$$ Where $m$ is mass of ball, $I$ is inertial moment of ball, $v$ is translational (linear) speed and $\omega$ is angular speed. For rolling without slipping the following condition also holds: $$v=\omega R$$ The ball will have ...

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The force balance is real - but you may have to take into account acceleration of mass along the way. Acceleration gives rise to an additional force - if I push onto a 1 kg mass with a force of 10 N, it will accelerate at 10 m/s/s. But if I put a second mass in front of the first, their total mass is 2 kg and when I push with a force of 10 N, they will ...

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