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3

This is a "famous" problem for showing what different things "average" can mean. If "the first half" means half of the time, then your distance will be $(v_1+v_2)*t/2$, and therefore the average velocity just the arithmetical mean $(v_1+v_2)/2$. If "the first half" means half of the distance, then your time will be $\frac{s/2}{v_1} + \frac{s/2}{v_2}$, ...


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I think you should take the sample velocities and divide them by the respective times after minusing the previous velocities to obtain the accelerations. If the time interval between calculating the discrete sample velocities are too small, then the above-got values may be taken as instantaneous acceleration. Now plot these over graph wrt time. Now here we ...


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No. Trivial counterexample would be if the origin, o, is just below the tangent point for v1, directly below it. In that case, beta clearly can be 90 degrees or more, even if v1 and v2 are actually in the same direction!


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If you draw similar triangles, then you'll find that $r_A/r_B = v_y/v_x$, and so the product $r_A v_x$ is equal to $r_B v_y$. Try drawing a line from the tip of your lower $\vec{v}$ vector to the tip of your lower $v_y$ component to see this.


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Do u really think that velocity is constant?i think in this equation nothing is constant.if tension increases per unit mass decreases and it may make change in velocity or not if the ratio remains the same.further tension depends on some variables such as intermolecular force,elasticity etc


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The tension of the string is a constant, if there is no vibration on the string. A wave is produced on the string when you give an unbalanced force on the string which varies the original tension of the string. The velocity of the wave now depends on the value of the tension. The given equation is valid only for small amplitude vibrations. The tension is ...


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When deriving the wave equation we assume the horizontal component of the tension in the string is constant and equal to $T$ (the tension when the string is at rest). To calculate the tension in the string let's start with the wave then zoom in to a small segment of it. If we take a segment small enough that we can consider it as a straight line, then the ...


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A fairly simple way of treating the data is to present them as a histogram: Each data point (here 5 data points) is the quotient of the distance moved in that interval, say $\Delta y$, by the time interval $\Delta t$ and is the average velocity during that time interval: $$v_i=\frac{\Delta y_i}{\Delta t_i}$$ Where $i$ indicates interval number $i$. For ...


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Let two orthonormal systems $Oxyz$, $O'x'y'z'$ with a general motion (translational plus rotational) between each other and a point particle $\rm P$, see Figure. Symbol Conventions : 1.The vectors for position $\mathbf{R}$, velocity $\mathbf{U}$ and acceleration $\mathbf{A}$ of a particle with respect to $Oxyz$ expressed by coordinates of this same ...


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Harmonic motion is sinusoidal: $x(t)=x_0\sin(ωt)=x_0\sin(2πt/τ)$. The argument of the sine function is called phase, here an increasing funtion of time: $x(t)=x_0\sin(φ(t))$ with $φ(t)=ωt=2πt/τ$. Hence, a phase difference $δφ$ corresponds to a time difference $δt=δφ/ω=δφ\,τ/2π$. With $δφ=π/2$ you find $δt=τ/4$. Remember: $1\ \text{period} ↔ 2π ↔ τ$


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Before valve-A is opened, the pressure in the tube is atm. Pressure since the top of the tube is opened. When the valve-A is opened, assumed instantaneously, the water will rush into the tube at: $$V = \sqrt{2gh}$$ where: $g$ = gravitation const (32.2 ft per sec per sec.), $h = 100\ \mathrm m$ = Height from bottom of tube to tank ...


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The real answer is quite complex; I think we should break it into a couple of different pieces. First - the static case. If you submerge an open pipe into water, the pressure inside and outside will be the same at a given height, and the water level inside the tube will settle at the same height as outside. If you add the effect of surface tension, it is ...


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To add to Dilithium Matrix's answer, to simplify, you can look at $v_1$ and $v_2$ as any vectors, not necessarily velocity vectors. We know that if $a = b$ and $b = c$, then $a = c$. Hence, to say that $\theta = \beta$ is to say that $\alpha = \beta$. You can then use geometry to show that there is only one other case where, for $O' \neq O$, $\alpha$ will ...


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As per the comments, I wasn't taking into account the relativistic addition of velocities, which is becomes relevant when designing scenarios with such high velocities. So for a observer in the point specified in my argument, the fastest objects (object #1 million, object #999.999, ...) would appear to have velocities close to light speed, but they would ...


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Hint: Since the car begins from rest and keeps accelerating, the equation $d=vt$ is no longer valid, because velocity is not constant. Note that you can, however, say that $d=v_0 t + {}^1/_2 a t^2 = {}^1/_2 a t^2$ and the acceleration will be the same for both time intervals. Hope that gives you a nudge in the right direction! Also, these equations may be ...


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In your solution you have found an acceleration and then used a constant acceleration kinematic equation. In terms of energy think about what $\int F_x\;dx$ might be equal to. PS Multiply your constant acceleration kinematic equation right through by $\frac 1 2 m$ and you might notice something interesting?


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As you said, the next derivative of the velocity with respect to time is the acceleration. And the acceleration could in principle have a step somewhere due to a force starting to act on the object.


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All you have shown is that the instantaneous velocity of the particle at point $D$ is can be equated to the translation of an arbitrary axis together with a rotation about that axis. The statement "while $\vec \omega$ is independent from the chosen rotation axis, $v$ depends on it" refers to all the particles in the body. So the all have the same ...


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I think that what you are asking is the position, velocity and acceleration of a space probe relative to the ever changing positions of the components which make up the Solar system? In the article cited below it describes the coordinate system as follows "Calculation of the trajectory of a space probe requires the use of an inertial coordinate system as ...



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