# Tag Info

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The indigenous activist will travel in a circular arc of radius 10m. Using this image: And this formula from Wolfram: $r$ $=$ $1/2$ $\sqrt{4R^2 - a^2}$ Where $a$ $=$ $5m$ and $h +r = R$ You can find $h$, the height dropped and hence the change in gravitational potential energy.

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The hypotenuse of the right triangle formed by the vine and the gorge is 11.18 meters. In order to drop from the end of the vine on to the far edge of the gorge, he will have to propel the vine to carry him upward, against the Earth's gravitational acceleration, so that his ending position is at the far corner of a new right triangle with base of at least 5 ...

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Lets say,we draw a straight line from a point A to a point B. if a body traveling under constant acceleration travels along that line from A to B.Now, if we divide that line into segments at equal intervals of time, it can be seen that length of each segments increases with the interval count in a linear way. Now draw another line parallel to AB say CD (just ...

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My take... First, consider constant velocity $v$, then by daily intuition, the average velocity $\bar v=v$ (because its the same $v$ throughout, thus the average cannot be anything other than $v$) Now consider acceleration, and the initial velocity $v_0$, Then the final velocity $v$ must be greater than $v_0$ The average $\bar v$, which is some kind of ...

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How about this: $v(t)$ is a line. This should be an OK assumption. So it's of the form $mt+b$. If you consider the interval $t_1$ through $t_2$ and draw a horizontal line of height $\frac{1}{2}(v(t_1)+v(t_2))=m \frac{1}{2}(t_1+t_2)+b$, you'll see that exactly half of the velocity line lies above the horizontal line, and exactly half lies below. Therefore ...

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Yes it is possible, though highly unlikely with random position and velocities. The probability goes up with each of the following: They are in a closed space where they bounce off from the edge. Initial direction is towards the general direction of the other particles. If you consider any point on the object instead of only the center of mass. You can ...

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Rotate the digram so the line connecting the circles is horizontal at the moment that they touch - since you know dx and dy, you just take the arc tangent. Now you move the frame of reference so the point where the two balls meet is stationary. The actual speed of the center of mass is just the vector mean of the velocities of the two balls (if they have ...

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So I came up with a graphical solution to this kind of problem. It might help you understand the process of collisions, without giving you a direct answer. Consider an Cartesian coordinate system xy for measuring momentum. Draw the initial momentum vectors $\vec{A} = m_A \vec{v}_A$ and $\vec{B} = m_B \vec{v}_B$. Draw a circle with the two vectors as ...

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Friction is awfully complicated because you're trying to model a system with infinite microscopic interaction with few macroscopic variables, It ought go haywire theoretically and hence all those "empiric" and "adhoc" stuff you have already found out. It's not possible to find a "unified theory of friction"(Well formally we have one, that is QED). That's why ...

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This problem requires a free-body diagram, where there is a friction force pushing the car forward, and air resistance and rolling resistance resisting the forward motion of the car. If there is a net force on the car (a non-zero force remaining after adding up all of the propulsive and retarding forces) the car will accelerate, either in the forward ...

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This type of problem can be simplified if you use the frame of reference of the elevator. Now the bolt falls from rest, chasing the "starting point" which starts a distance $h$ below and moves down at a constant $6$ m/s. The bolt catches up in $3$ seconds. Same problem, but the equations are simpler...

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This is an excellent question. It is pedantic to whine about non differentiability ,however there is in fact a point to be made on that topic . We seem to be conflating the derivative dy/dx with the time derivative dx/dt(x is the position here). One may have a differentiable path but still the instantaneous velocity can remain undefined. Whether the converse ...

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No, it's not possible, because one of the underlying assumptions of kinematics is that all paths are at least twice differentiable. Before you complain about this requirement, remember that physics is about building models that can be used to describe and predict measurements. Measurements always have some amount of uncertainty, and even if you suppose that ...

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The length along any segment of the Koch snowflake is infinite. It has finite area but infinite perimeter. So, for a particle to move from one place on the snowflake to another it would have to travel an infinite distance. This is why differentiability is important.

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I am assuming you have 2 particles facing each other, and that they are approaching each other ? First, as mentioned elsewhere on this page, "..a particle moving at the speed of light does not experience time, and thus is unable to make any measurements." Instead, let's change the particle #1 that you are sitting on to having a specific velocity that is ...

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Starting from position measurements $x[i]$ taken at times $t[i]$, the total distance travelled is given by $$d[i] = \sum_{j=0}^{i-1}\left|x[j+1] - x[j]\right|$$ or, equivalently $$d[i] = \sum_{j=0}^{i-1}\left|\Delta x[j]\right|$$ where $\Delta x[j] = x[j+1] - x[j]$. Note the absolute values in the sum formulas.

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Your situation is dynamic in that the ball accelerates under the force of gravity, initially at rest but then reaching a steady state, terminal velocity, similar to what a skydiver experiences when jumping from an aircraft. The terminal velocity is reached when the force of gravity is in balance (equilibrated) with the viscous drag force imposed by the flow ...

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The velocity is downward, and the acceleration is downward. Whatever direction you choose, if you start with a velocity of zero the sign of both will be the same (if you throw the feather down, it will decelerate - so the acceleration will the "up". I don't think that is intended here). Whether the floor or the hand is zero in the coordinate system doesn't ...

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Change of velocity is definitely -8m/s, according to vector summation. Magnitude of change of velocity is 8m/s. '-' sign indicates that the change is taking in the other direction,i.e. in the direction opposite to the initial direction. 4-4=0 is the change in speed , not velocity

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If you define your axes in such a way that the runner has a speed of 4m/s in the y direction in point A, the the velocity in point A is: $V_A = 4 \, \text{m/s} \, \hat{y} + 0 \, \text{m/s} \, \hat{x}$ Your runner must have a speed of -4m/s in the y direction in point B, you get the velocity: $V_B = -4 \, \text{m/s} \, \hat{y} + 0 \, \text{m/s} \, \hat{x}$ ...

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In general, the position is the integral of the velocity - this is the essence of the equation you wrote down (without the "terminal velocity" part which makes no sense). If, at a certain time, you change the velocity (direction), you can just treat that point as the start of a new trajectory: the initial velocity will be "the velocity of the turned ...

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The classical equations of motion should do it for you. This one for example: $$v_{final}=v_{initial}+at$$ Plug in your maximum deceleration (the most negative $a$ possible) and you other values. You can easily find the time $t$ is will take to reach 0 m/s. If the value for the maximum deceleration is not constant, then those equations don't apply. ...

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However, I find that the net torque of the cylinder is 0 if I set the axis of rotation as the bottom of the cylinder (the contact point between the ground and the cylinder); my first question is, what accounts for this discrepancy? I suspect there should be a force about the center of mass, but I don't see what could be the source of such force. You're ...

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The force on a current carrying wire is given by: $$d\vec F= I d \vec l \times \vec B$$ Or for a uniform current and field and in scalar form: $$F=BIl$$ If you want to derive this from $$F=qvB$$ then you can note that in a time $t$ a total charge of $\frac{q}{L} vt$ passes through a point in your conductor. Since current is charge per unit time we have: ...

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Theoretically speaking, of course, if your true linear speed with respect to the true center of the universe was zero, you would be experiencing true time. Even on Earth, who's movement is what we base our time (e.i. seconds, days, years, etc) off of, we theoretically would be experiencing time dilation based on Lorentz, assuming that at a true zero ...

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The basic reason why your approach is too simple So the first problem for you to meditate on is: $\int dt~ r \ne r t,$ because $r = r(t)$ is going to be depending on time. For example, if something is purely falling up/down radially towards/away from the Earth, then the proper expression is instead:$$m \ddot r = -\frac{GM}{r^2},$$and the proper way to solve ...

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I was able to determine the user status(static, slow walking, fast walking) by calculating the variance. The Va in the research was not velocity. It was my mistake to interpret it as such. It was the variance of the euclidian norm of the accelerometer data. I decreased the accelerometer update interval to 0.1s and every second I took 10 of the values and ...

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Ok so there is one major problem! You have assumed that because your going at half the speed of light and that bob is going at half the speed of light in the other direction then, relative to you bob goes at the speed of light, but you don't add up velocities in relativity. If your going at velocity $v$ and bob goes at velocity $-u$ then, its velocity ...

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Air drag is given by $$F_d = \frac12 \rho v^2 A C_D$$ where $\rho$ is the density of air (variable, depending on temperature etc - 1.2 kg/m$^3$ is a reasonable approximation), $v$ is the velocity, $A$ is the projected area, and $C_D$ is the drag factor - which is a function of shape and Reynolds number. 0.5 is an OK approximation but it depends on the ...

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According to WolframAlpha, $100\text{lbs}$ of ice has a volume $V=0.04536 \text{m}^3$ and mass $m=45.36\text{kg}$. We can find the radius of the ice ball: $$V=\frac{4}{3}\pi r^3 \rightarrow r = \sqrt[3]{\frac{3V}{4\pi}} = 0.221\text{m} = 22.1\text{cm}$$ The force of air resistance is given by the equation: $$f_\text{drag}=-\frac{1}{2}C\rho Av^2$$ Where ...

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This is really just a simplified version of Timaeus' answer, so please accept his answer not mine. Anyhow, you're quite correct that the ball gains energy, but that energy doens't appear from nowhere. in any (inertial) frame total energy is always conserved. What you are seeing is some of the kinetic energy of me and the room being transferred to the ball. ...

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Try arresting the balls motion while standing on a path of ice. What happens is the ball and that floor come to a relative rest. And any force exerted on the ball to slow it down or speed it up has an equal and opposite force exerted on the thing accelerating it. You can analyze it in any frame and get a correct description (though kinetic energy depends ...

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