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Let the velocity we wish to find be v1, which is the initial velocity Therefore initial energy is (1/2)m(v1)^2 ----------(1) Let v2 be the velocity at top Therefore final energy is (1/2)m(v2)^2 +mg(2r) -------------------------------(2) Now we take the force equations when the car reaches the top mg-N=m(v2)^2/r --------------------------------------(3) ...


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The basic question is how the speed of a rocket change. Since you are always expelling mass at a fixed velocity, with a given momentum, you must also gain that momentum yourself, so that it is conserved. Now assume that by some reason, even though you are expelling mass, your mass doesn't change. You are throwing mass with a fixed velocity, so your own ...


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You asked for an intuitive answer. A rocket accelerated by burning fuel and expelling the combustion products at high velocity. Conservation of momentum says that if you expel the same mass faster, you will get greater acceleration. This gives rise to the proportionality with $v_{cx}$. As for the logarithmic part: if you imagine two rockets of mass $m$ ...


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The equation is basically telling you that the change of velocity of the rocket, $v-v_0$, is proportional to the expulsion velocity of the fuel $v_{ex}$. But it is not quite equal, since as the rocket is burning fuel, it is getting lighter. This is why you also take into account the rate of change of the mass, $\ln(m_0/m)$. So, how do we derive this? ...


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I assume the question asks for the minimum initial speed needed for a car on a horizontal track to enter a loop and remain in contact with it throughout. Assume the speed at the top of the loop is the minimum needed. That means the centripetal force is being supplied by gravity alone (no normal force at that instant). We can write this as v^2/r = g. The ...


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The initial energy is the kinetic energy and the gravitational potential added together. The final energy, when the rock hits the ground, is entirely kinetic and equal to 450J. Energy is always conserved. So the total energy of the rock is 450J through out the problem. Throwing the rock in a different direction does not change the initial kinetic energy ...


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Your answer is correct because when looking at change in motion of an object, you look for the "net force" which is the sum of all forces in a scenario. Examples of the net force being 0 are: The car is parked and Jared is standing next to it, not pushing on the car. Jared exerts 0N on the car which is exerting 0N on Jared. No motion happens Jared turns ...


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For Jared to push the car home he must cause some acceleration to start the car moving from rest. This is an application of Newton's 2nd law $\vec{F}_{total}=m\vec{a}$. The forces you need to consider are the force Jared can supply and the frictional force on the car. For Jared to initially cause the car to accelerate the force he applies must be greater ...


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Assuming the total force is $10~\text{N}$, the $7~\text{kg}$ mass is accelerated by $1.0~\text{m/s^2}$, so it is experiencing $7~\text{N}$ force in x-direction. Therefore, $F_x = .7F$. The angle is roughly $45^\circ$.


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Suppose that as it slows down the projectile has a mass $m$, velocity profile $v(t)$ and cross-sectional area in contact with the target $A(t)$. Then the force is $ m ~ dv/dt$ and the pressure is $ (m ~ dv/dt) / A$. We can't answer with much more specificity than that, but if the velocity profile is a straight line from $(0, V)$ to $(T, 0)$ taking time $T$ ...


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Resolve the force vector into components. The horizontal component is equal to $\vec{F_x} = 5.2\cos{60^\circ} \hat{i}$. Then you can find the horizontal acceleration of the body by inserting the mass in Netwon's Second law ie. $$ \vec{F_x} = 2.6 \cdot \vec{a} \implies \vec{a} = \dfrac{5.2 \cdot 0.5}{2.6} \implies \vec{a} = 1~\text{m/s^2}$$ . Thus, the body ...


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Your $2.0 \frac{m}{s^2}$ is the acceleration in the direction of the force, at $60°$ angle with the x-axis. You need to find the component of the acceleration in the direction of the x-axis. If your acceleration is completely in the y-direction, your object will have $0$ acceleration in x-direction. If it is completely in the x-direction, it will have those ...


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Newton actually calculated, using a lot of obscure geometry and limiting concepts, the orbits that various force forms would generate. One of those forms was the inverse square force. If you want to know what the others were, find a copy of the Principia and wade through it. The result wasn't published until Edmund Halley asked Newton if he knew the nature ...


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Weight is a subtle concept, because we are so used to it we don't even notice it anymore. You'll find surprising how badly some students grasp the concept of weight, despite the fact they are firmly sitting on a chair in the very same moment. Anyway, weight is the force an object experiences when inside a gravitational field. That means that me, you, ...


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John von Neumann researched self reproducing automata. He settled on 2-D (tiled) systems to show that such systems could replicate themselves. The problem is not informational; it is practical. His reproducing automata formed the basis of Conway's "Life" programs.


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While both are forces, weight is generally specific to any sum of forces you feel reciprocated as a normal force (or tension). I can feel heavy in a centrifuge because of the centrifugal force. When I find myself sitting in my chair (safely away from centrifuges) the astronauts on the International Space Station and I are both are subject to the force of ...


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I want to take another tack than that of the other answers. This will be one big handwave rather than a rigorous mathematical argument, but I hope it gets the idea across intuitively. First off, as I noted in a comment, and as hft notes, you are using "v" to mean both "velocity as a function of time" and "velocity as a function of position". That's ...


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Variations of this problem show up all the time. If you start with the spring "locked" and the spheres charged, then release the spring, it will expand to the new length and when it gets there the spheres will have a velocity - essentially you have a simple harmonic oscillator and the point of (new) equilibrium is the point where the oscillator moved ...


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The rotational energy of a body is given by: $$ E = \tfrac{1}{2}I\omega^2 $$ where $I$ is the moment of inertia and $\omega$ is the angular velocity. For a uniform sphere the moment of inertia is related to the mass of the sphere, $m$, and the radius of the sphere, $r$, by: $$ I = \frac{2}{5}mr^2 $$ You already have the mass, and you can Google for the ...


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Yes , the normal to the surface is the direction of reaction force. And the direction doesnt depend on the material of the object . But note that if friction is considered , direction of net reaction force changes


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Initially a small force needs to be applied to get the desired velocity but this small force is neglected and we start our observation after the body acquires the velocity and the force is removed .


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Note that when you apply chain rule , you assume dx not to be zero . This will clear it up for you .


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You can apply chain rule if $v$ is differentiable wrt $x$ and $x$ is differentiable wrt $t$. I think there are no other conditions,as this post on MathSE seems to say, http://math.stackexchange.com/questions/688152/necessary-conditions-for-the-chain-rule-of-differentiation-to-be-valid#= and this condition is not always available. When $v=0$,make sure ...


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No, it doesn't imply that $a = 0$. If, at some value $t = t_0$, the acceleration is non-zero while the velocity is zero, the position function is either a minimum or maximum. That is, $x(t)$ is stationary there: $$x(t_0 + dt) = x(t_0)$$ which means that at $t = t_0$ $$\frac{dx}{d\dot x} = \frac{dx}{dv} = 0$$ thus $\frac{dv}{dx}$ is undefined at $t = ...


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The correct thing to say would be that "if v=0 and dv/dx is finite then a=0". A simple example, to help illustrate what's going on, is the well known case of constant acceleration "-g" near the earth's surface. In this example, we consider "x" to be the height above the ground, and assume the initial x is zero. In this case $$ x=-\frac{gt^2}{2}+v_0t $$ $$ ...


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If we have a graph on a pc screen of velocity versus time OK, the plot is of velocity versus time. I don't quite understand what is meant here because when I tried this machine and when I move away from the sensor, the cursor on the screen draws a line with positive slope and when I move toward the sensor it draws a negative slope. But the ...


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For a sensor that detects motion I would say what you described is the more intuitive convention to use. Imagine the sensor measured distance. Then when you are close to the sensor it will measure a small distance, and when you are far it will measure a large distance. Say you start at 1 foot from the sensor and move 10 feet in 3 seconds away from the ...


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I have an answer, but I was hoping to see confirmation from another source. I used the centripetal force formula to come up with: $$ v = \sqrt{Gm/r-ar} $$ Where v is the tangential velocity of the orbit and a is the radial acceleration away from Earth. Is this right?


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Since the force is radial, you are not changing the angular momentum, but you are adding potential energy: this tells you what must happen to the tangential velocity (decreases) and radial velocity (increases). I will leave it up to you to figure out by how much.


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Because it is a perfectly elastic collision the kinetic energy and the momentum are conserved. So you have two equations for two unknowns which are the final velocity of the football player and his mass: $$ m_f v_f^0+m_r v_r^0=m_f v_f^1+m_r v_r^1 $$ $$ \frac{m_f (v_f^0)^2}{2}+\frac{m_r (v_r^0)^2}{2}=\frac{m_f (v_f^1)^2}{2}+\frac{m_r (v_r^1)^2}{2} $$ and ...


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In a perfectly elastic collision, the final momentum of the system should be equal to the initial momentum of the system. It seems to be set up correctly, so I would say that you would need additional information for this.


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Use conservation of momentum, which tells you that the total momentum (the sum of the momenta of the two particles) before and after collision must be the same. Also note that the momentum is a function of the vector velocity, which means that you can make two independent analyses, one on the $x$-axis, and one on the $y$-axis. Both should respect ...


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After the mass exits the pipe, the tube will start to rotate from the recoil. At $t=0$, there is zero angular momentum, $L=0$. Let's take it that the pipe rotates about its centre-of-mass, and use that point as the origin from which to calculate the angular momentum. At time $t+dt$, the puff of gas has angular momentum $L_{gas} = dm \times l/2 \times v0$. ...


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As far as I know, when a wheel rolls w/o slipping, it always has friction in the opposite direction of its movement. Here you go wrong . In pure rolling (that is no slipping ) the bottom point is at rest wrt ground . So there is no kinetic friction acting on it as there is no relative motion .But static friction acts . In your question gravity ...


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Friction acts on objects at rest too . The definition meant that if there is relative motion between two objects then friction will act as a resistance between them . If you find two objects at rest even when an external force is applied on it then it means friction is acting on them . Had there been no friction there would have been relative motion and the ...


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The steps in the OP's suggested solution are deeply flawed. First: you cannot take the absolute values of the two sides of a vector equation. Momentum has both size and direction, and both must be taken into account when doing an addition. Would you examine your bank statement for the month, while treating the deposits and withdrawals differently. Is a ...


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In lower gravity, you could expect to swim faster I am not answering the other questions as I do not have much more to say which is not already said in other answers. But I do disagree with their conclusion that swimming would be the same. Regarding swimming, one would need a better understanding of swimming motion to decide how much effect can be expected ...


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But if I take the absolute value of both sides, and drop summation It might be dangerous just to "drop" summation. You have to include all particles moving before and all moving after collision. In your case of only a ball and a wall it would reduce to: $$\sum m v_{1x}=\sum m v_{2x} \Rightarrow\\ m_{ball} v_{1x,ball}+m_{wall} v_{1x,wall}=m_{ball} ...


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This was a tough question to crack and I wanted the Stack Exchange to carry the answer, as the community is a great resource. If you dont realize that her speed is relative to the plank, you'll be wondering what kind of problem this is. Answer: There is no external force on the system, so the sum of momentums stays = 0. And velocity of the girl relative ...


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Considering both the discs as the system , we can conserve angular momentum about their collinear axis of rotation . The torque due to friction will decrease the angular velocity of the disc having more angular momentum (before the collision ) while the torque will increase angular velocity of the one which had lesser initial angular momentum . I am assuming ...


0

I've read a little bit on the science of a space elevator and it's a surprisingly difficult problem. To have a working space elevator, it would need to be at least to the Geosynchronous orbit, 22,000 miles up, probobly a bit beyond that for buoyancy. The highest balloon is some 25 miles - so that's less than 1/10th of 1% of the distance. The strongest ...


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"Equation that is all over the internet"... You started at http://thatsmaths.com/2014/06/26/balancing-a-pencil/ and from there, you linked to http://arxiv.org/pdf/1406.1125v1.pdf which was the source for the former. In the third paragraph of that paper, it states We model the pencil as an inverted simple pendulum with a bob of mass m at one end of ...


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If the cable of the "elevator" is not connected to a point on earth, then the satellite must be in a geostationary orbit (or it will float away); this implies that if you now attach something to the platform (increasing the pull on the cable) you will pull the satellite down to earth. And as @lionelbrits pointed out, the pulling part of a space elevator ...


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What I cannot understand is, why acceleration, a=lθ¨ and not lθ¨/2? The equation you wrote doesn't mention anything about the linear acceleration. Is the center of mass located at its top and not the center? Or is there something else I am missing? The center of mass of the pencil is in the middle, not the top. There is likely something else ...


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The simplest way to look at this is to consider separately the horizontal and vertical velocity/position. For a projectile launched at angle $\theta$ and velocity $v$, the components are: Horizontal velocity $$v_h = v\cos\theta$$ Vertical velocity $$v_v = v\sin\theta$$ The position at time $t$ is then given by $$(x, y) = (v_h\cdot t, v_v \cdot t - ...


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logically in the frame of centre of mass the accn of body is zero so momentum is conserved and as mass has not changed initial velocity is equal to final velocity in frame of centre of mass


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In addition to Pentane's solution, this can be done as a projectile problem as well. The following is a basic kinematics equation $$v_f^2 = v_i^2 + 2 a \Delta x$$ At the top of your jump $v_f=0$, so you can solve for $v_i$ and get the answer that way.


0

Swimming would be nearly identical to a 1g planet, other than the splash being bigger. The forces involved in swimming are largely horizontal, so as long as there is some gravity to keep the water where it belongs you are acting against the viscosity of the water rather than the weight of the water. Might be a problem at very low g as you would splash away ...


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Start with the two pertinent conservation laws for elastic collisions: kinetic energy and momentum. Remember that momentum is a vector. In the center of mass frame, the total momentum is zero. That will get you started. Do the work for two particles first. As an aside you should try to show the total momentum is zero in the CoM frame by example by taking ...


0

You can use conservation of energy: the kinetic energy when they land has to be the same as the kinetic energy when they left the ground. The kinetic energy when they land is also equal to the potential energy when the person is 60 cm off of the ground. So you can use: $$mgh=\frac{1}{2}mv^2$$ and the mass cancels out leaving: $$gh=\frac{1}{2}v^2$$



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