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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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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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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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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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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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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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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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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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In Galilean relativity (pretty sure that is what you mean/need), you just add up the velocities. So $ V_{p_1} =v/2+x$ and $V_{p_2}=v/2+y=v/2-x$


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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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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 ...


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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 = ...


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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 ...


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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 ...


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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 ...


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The small object exerts a force in the opposite direction to the normal force on the cart


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You should go and find another professor. For the left hand mass because it is in equilibrium T-mg=0 and there is a similar equation for the right hand mass. For the gauge T-T=0. So the gauge is reading the value of T.


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Your error is putting $g = -9.8$ in your evaluation of $m\frac {v^2}{ r}+mg$ because you already have indicated the direction of the gravitational force with a negative sign in $ \vec{F_{net}} = \vec{F_T} - \vec{F_{grav}} $ Put another way $ \vec{F_{net}} = \vec{F_T} + \vec{F_{grav}} = \vec {F_T}+ m (-\vec g)$


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This equation holds whenever there is constant acceleration. Here are 2 ways of deriving that equation, which I hope help you understand it. Energy conservation The change in kinetic energy must be equal to the work done on the particle. $$ \frac{1}{2}m v_A^2 - \frac{1}{2}mv_B^2 = \int F\cdot dx $$ For a constant force and mass $\int F\cdot dx = F (x_A - ...


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You are right to be skeptical of your answer. The technical specs show you should get something on the order of 100 m$^2$ . Your problem lies in basic trig: the force of the pressure is normal to the surface, and the surface is tilted by 5° - this means that the vertical component of force will be $\cos 5°$ times the normal force - you used the $\sin$ which ...


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 ...


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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 ...


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This assumes a smooth surface collision. The component of velocity (momentum) along the surface of the wall cannot change because there is no friction and hence no forces along that direction. Because the mass of the wall is assumed to be much greater that the mass of the ball and the collision is assumed to be elastic the normal component of velocity of the ...


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Exactly. In a 2D problem, it's usually a good idea to break the components into two dimensions based on the environment. In this case, the wall makes the best split, let's say that x is the direction of motion along the wall while y is perpendicular. In this simple situation, the force on the ball can only act in the Y direction. Which has the following ...


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The max resultant force for these two forces would occur if they were both acting along the same line in the same direction. In that case they add algebraically. 3+4 = 7. Any other configuration of the forces gives an answer less than that.


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let's call the forces and their norms : $\Vert \vec{u}\Vert = u = 3 $ , $\Vert \vec{v}\Vert = v = 4 $ and $\vec {F} $ their sum , of norm F . The 2 vectors have an angle $\alpha$ between them. the Al Kashi Theorem says : ${F^2} = u^2 + v^2 + 2 . u . v . cos(\alpha)$ We got the maximum with $cos(\alpha)=1$ and the minimum with $cos(\alpha)=-1$ ...


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It is true that the heavier plane needs a greater lift and this is seen in practice. If two planes at equal altitude loose power at the same time and one weighs more than the other they will be able to glide the ...... same distance! One of them descends faster than the other but it glides forward faster to generates more lift. It seems odd, but one ...


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I came up with a solution seeing John Rennie's comment. The centripetal force, $\vec F= -F \hat r $ so infinitesimal work done by centripetal force, $$dW=\vec F.d \vec r= -F \hat r.d\vec r$$ but, $\hat r⊥d \vec r$ so $$dW=0$$ is this correct ?


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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 ...



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