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To begin with ask yourself two questions: Has the device been re-calibrated in the middle of data taking? Is the calibration known to drift over time similar to the length of the data taking? If both of the answers are "no" then you can reasonably assume that the calibration effect is the same on each and every data point. So the mean is off by that ...


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Just visualize the upper beaker as a tube connected to the siphon tube(both tubes having different area of cross section)


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In this case, it is allowed. As a thought experiment, you can replace the top bucket with the siphon hose by an S-shaped tube that has a gradual change in diameter, like this: _ / \ |www| | | \www| | | \ww| | | \w| | | \|__/ | | | With this type of analysis, it is more ...


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Accelerations being equal doesn't necessarily mean that the velocities are equal, or vice versa. For example, your two cars could have the same acceleration, but if one starts before the other, the one that got going earlier wlil obviously be moving faster. An even simpler example, if one car is standing still and the other one is moving at constant speed, ...


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there are ways to find the velocity- you can look at the distance and time graph of the motion of the ball and the slope at a point of time will give ya the velocity. If you are looking at the acceleration, well the initial velocity must be zero I guess because it starts at rest...


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Your equation for the $\mathbf{v_{AB}}$, the velocity of $\mathcal{A}$ relative to $\mathcal{B}$ with the following formula is correct and general for all 'real' values of $v_{X_{i}}$ where $X=A, B$ and $i=x, y, z$. ...


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Calm down. aren't there an infinite number of distances. This Statement is necessarily false. Distance is an scalar and is unique with a magnitude. It is true that that the possible distances are infinite but at a particular time (say t = 10 min) there will be an unique value of distance. The answer to what term do we use to refer to the rate of ...


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You are overthinking this. Speed is rate of change of distance. Velocity is rate of change of displacement. Because distance and displacement are not necessarily the same, speed and distance are not necessarily the same either. Yes, you are correct - many different distances could correspond to the same displacement, so one particular velocity could ...


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Use the work-energy theorem to find the speed of the fragment that you were travelling in after the collision. Find the net momentum North and the net momentum West after the collision. Find the magnitude of the momentum and its direction after the collision which will be the magnitude of the momentum and the direction of the car and occupants of mass 1600 ...


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If you draw a velocity-time graph you will see that you do not have enough information to find the initial and final velocity. The gradient of the graph is fixed because it is the acceleration. The distance is the area under the velocity-time graph. As you will see from the graph you can draw an infinite number of trapeziums (or triangles) which satisfy ...


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Assuming constant acceleration You have to know the time also. If you know the distance traveled $s$ after time $t$ then you can write $$ s = v_0 t + \frac{1}{2} a t^2 $$ and solve for the initial velocity $$v_0 = \frac{s}{t} - \frac{a t}{2} $$ Once the initial velocity is known, then the final velocity is $$ v_1 = v_0 + a t $$


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No, that's not possible. Unless the body is starting or ending at rest in which case $v_f$ or $v_i$ would be zero and you could substitute in $0$ in the eq: $v_f^2 = v_i^2+2aS$ Intuitively this makes sense. If the acceleration and distance for which the body accelerates are known, you can only determine $v_f^2- v_i^2$ You don't know whether $V_i = 0m/s$ and ...


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Under constant acceleration, the relationships between velocity, acceleration, distance and time are: $$\begin{align}v_t &= v_0 + a_t\\ x_t &= x_0 + v_0 t + \frac12 a t^2\end{align}$$ When a and d ($x_t$) are given, you are left with two equations and three unknowns: $v_0$, $v_t$ and $t$. This means you cannot come up with a numerical (unique) ...


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(Source of image: Mohsin Khan, http://cslearners.blogspot.com/2009/08/equation-of-motion.html) Here they are! All the formulas. Sorry to say, you cannot find anything if you have only acceleration and distance. Think like this, Say you have an object that has an acceleration of 2 m/s^2. and if i say that if travels a distance of 2 meters. It can travel ...


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If I understand correctly, the speed of the vehicle as it passes a particular point must be one of the values in a list. The list is different for each point. The overall time is minimized by choosing the maximum speeds possible at each "checkpoint", subject to the constraints. I think what you might be missing is that, if the starting speed on a section ...


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'valerio92' is correct and for: $$\frac{d}{dt} u(t) + A \ u(t)^{3/2} + B = 0$$ ... there probably are no analytical solutions (wolfram alpha's DSolve for instance yielded nothing). But $F_d=C_d A \frac{\rho}{2}v^2$ is not the only model for drag forces. At low speeds: $$F_d=kv$$ ... can be used, so linear dependence of the drag force on speed. The ...


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You will obtain a first-order nonlinear differential equation: $$P=P-P_d = P-D v^3 = \frac{m}{2} \frac{d}{dt}(v^2) $$ $$\to \frac{d}{dt} (v^2) + \frac{2D}{m} v^3 - \frac{2P}{m} = 0$$ If we define $u(t)=v(t)^2$ , $2D/m = A$ and $-2P/m = B$, we obtain $$\frac{d}{dt} u(t) + A \ u(t)^{3/2} + B = 0$$ When $A$ is small this reduces to the result you obtained ...


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This is a quite subtle problem. You have to be careful about three different situations. A ball can be thrown with velocity (relative to the ground): a) $v_0-v_e$. b) $v(t)-v_e$, where $v(t)$ is the velocity of the car just after the ball is thrown. c) $v(t)-v_e$, where $v(t)$ is the velocity of the car just before the ball is thrown. You actually stated ...


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All of your math is correct, but your result conflicts with our intuition as well as the reality of how rockets work. Here is why: You assume that "The ball will be moving at $v_\mathrm{ball}=v_0-v_e$". This approximation is valid only in the limit where $m$ is much smaller than $M$. (i.e. the limit of a continuous stream of tiny balls, like a normal ...


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The answer is "it depends". If you have a very thin and light "rod" (for example, a piece of fishing wire), the material from which that is made has a shear modulus, and it is in principle possible to have a piece of fishing wire vibrate without being held in tension. However, if you pull both ends tight, like the string of a guitar, then waves will travel ...


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The formula is $$v = \varphi\cdot r/t$$ Where v is the velocity, φ the traversed angle in radians, r the distance and t the time, if the object is moving in a transversal direction. If it is approaching or receding you have to split the radial and transversal components with Pythagoras. In your example I get around 1140 ft/sec.


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From the definition of work $$W = \int dx F$$ and $$P = \frac{dW}{dt}$$ you can see how we can arrive at $$P = F \frac{dx}{dt} = F v$$ (when considering only the absolute value). To understand it intuitively, imagine the case of a frictionless system in which the car can move at a certain speed without any opposing force. The power required to keep it at ...


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This is a standard projectile motion problem with the complication that the projectile collides elastically with vertical walls. Assuming that the only force exerted on the projectile by the walls is in a direction which is normal to the walls then: the vertical velocity does not change, and the horizontal velocity reverses direction but the magnitude ...


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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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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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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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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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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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