# Tag Info

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This came to my mind after reading some introduction to maximum entropy probability distributions. Independence can be derived from the following four assumptions: (1) average momentum of particles inside the box is fixed at 0 (2) average kinetic energy of particles inside the box is fixed (3) the gas velocity distribution must be maximum entropy ...

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Break the problem into two parts: The x and y component of the trajectory. The x direction has no acceleration hence $$x=x_{0}+v_{x0}t$$ and the y direction has gravity as its acceleration hence $$y=y_{0}+v_{y0}t+1/2gt^2$$ Where $v_{x0}=v_{0}\cos\theta$ and $v_{y0}=v_{0}\sin\theta$ You can get the total time it is in the air by setting $y=y_{0}$ (the ...

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You've done the "most difficult" which is to get as many independant equations as needed to solve for the unknowns. The first one flows from the fact that the vertical component of the swimmer's velocity is the only one to make him reach the end of the river, and the second one from the fact that both the current and the horizontal component of the swimmer's ...

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I think I see it. The question should say "when the speeds are equal" not the velocities. The speeds of the two balls will not be the same at all points, but they are related, and there is one point (before the upper ball hits the corner) when the speeds are equal. In fact, the point at which the speeds are equal can be deduced from symmetry, but it's ...

1

The rear of the bus normally overhangs the back wheel further than the front of the bus overhangs the front wheel - especially on traditional yellow school buses. When the bus goes over a bump it pivots around the other set of wheels, the distance you move is the ratio of the distance between the wheels and the distance from your back seat to the front ...

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Someone posted the energy conservation way (if you have spring constant and distance stretched)so here is the Newtons laws way: If you know the mass of the object and the force on the object and the distance the has the force applied on it then you get the velocity as it leaves the slingshot (given newtons laws: \begin{align}F=ma \\\ F/m=a\end{align} ...

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I believe if you're modelling the slingshot as a spring, you can consider the potential energy stored within it when it's loaded is fully transferred to the kinetic energy of the projectile (neglecting the friction during the short acceleration etc.), and then it gets down to the ballistic of a freefalling object with air friction (see the drag coefficient ...

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One must distinguish between instantaneous velocity and average velocity.

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Let's take a look just from the point of view of someone reading the problem. First of all, we can say "our car has motion", because it's changing its position each second. Ok. So, how is its motion? Well, it is moving in 1 dimention, it is a linear movement. Then, we can say "our car has linear motion". Also, we can see our car's velocity is changing ...

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in v-t graph we see how velocity changes as time passes. if the graph is a straight line making acute angle with time-axis (x-axis), it means velocity is increasing at a constant rate. Now the problem here is, if i understood you correctly, is not thinking about it without using mathematics. In your question, the acceleration is constant and it is 10 meters ...

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The important thing is that all speeds seem to change: If you would look at the universe, in this case the two protons, from the perception of a man shrunk to the size of a proton, not only would the particles appear much faster, but so would the speed of light. So if you shrunk yourself to a trillionth of what you are now, one proton would have a diameter ...

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The car probably experiences a constant acceleration of $10{m/s^2}$. You can see from the chart that the velocity follows this as after every second the car is going $10{m/s^2}$ faster. However, this is clearly not the whole picture. We do not know the acceleration at 1.5 seconds, or 1.55 or 3.14. We can get confirmation that our acceleration model works ...

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You must consider the acceleration that is acting on the object. Let me give you an example, a ball falls from rest to the floor. This ball is 5m above the ground and the acceleration on the ball due to gravity is 10m/s^2. When the ball is let go, clearly the ball is not moving at 0m/s which is what it would be if there were no acceleration, it would just ...

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The velocity is constantly increasing due to a constant acceleration. Exactly at 1 s the velocity is 10 m/s, but this does not mean that velocity was at 10 m/s in preceding second. In fact, given the distance 5 m moved in this second, the average velocity in this second was 5 m/s. And this should make sense to you, because in this first second the velocity ...

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If I understand you correctly, your two points about apparent slowness of speeds is related to scale, and disappears when you quantify it using a common unit. ie: We think of 10m/s as relatively slow because the average human is 1.8 metres in height, and we can imagine that 10 metres per second, or 36 kilometer/hour as an achievable speed using a machine ...

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If you know the total force as a function of time, then you know the acceleration as a function of time also $$a(t) = \frac{\sum F(t)}{m}$$ Now you find the velocity and acceleration using direct integration $$v(t) = v_0 + \int a(t)\,{\rm d} t \\ x(t) = x_0 + \int v(t)\,{\rm d} t$$ If the forces are constant then you can convert the integral into ...

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You should consider Newton's 2nd Law: $\vec{F} = m\vec{a}$ and relate the forces acting on the object to accelerations. Once you know the accelerations for different periods of time you have broken down the problem into a simple kinematics one of an object moving with constant acceleration.

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Just use the free-fall equation. The time spent in falling from a height $h$ verifies this: $h - \frac{g}{2} t^2 = 0$ So you get: $g = \frac{2h}{t^2}$ Note that you have to determine the height $h$. How?, maybe you can estimate it thinking that the game's character is about 1.80 m. The $g$ you are obtaining here is the acceleration of the ...

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The two existing answers have both done the correct calculation, but both have forgotten to account for the Sun's gravity. A comet falling from the fringes of the Solar System is accelerated mainly by the Sun's gravity. We can see this from the expression for the potential energy at a distance $r$: $$V = -\frac{GMm}{r}$$ For the Sun $M = 1.9891 \times ... 1 The gravitational energy of the comet at infinity gets converted into kinetic energy of the comet. Calling$m$the mass of the comet,$M$the mass of the Earth,$r$the radius of the Earth we have: $$G\frac{mM}{r} = \frac{1}{2}m v^2$$ where$G$is the gravitation constant and$v$is the speed of the comet when it hits the surface. Thus:$\$ v = ...

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In both cases, the person who jumps has inertia of motion and hence will land in front wrt the earth's reference frame. It is the same effect which makes you jump forward and keep moving for a while when getting down from a slowing bus/train.

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My first guess is that if feels faster since your eyes are closer to the line zipping by. This the same effect that makes it look like you are going slower down the highway compared to distant buildings than to stuff at the edge of the road. The shorter line would have slightly less wind resistance, but that should be a insignificant effect in the scheme ...

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