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4

\begin{equation} \mathbf{v}\left(t\right)\equiv \dfrac{d\mathbf{r}}{dt}= \dot{\mathbf{r}} \tag{01} \end{equation} We'll use one upper dot for the 1st derivative with respect to $\:t\:$, for example \begin{equation} \dot{\mathbf{r}}\equiv \dfrac{d\mathbf{r}}{dt}\;, \quad \dot{\theta}\equiv \dfrac{d\theta}{dt} \tag{02} \end{equation} Now, let a system of ...


3

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 ...


2

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, ...


2

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 ...


2

(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 ...


2

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 ...


2

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 ...


2

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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Yes. It just means that the velocity is in a direction opposite the direction of your reference frame. If you make "down = positive" then $g$ would be positive, and so would the velocity.


1

Consider the picture below. In Cartesian coordinates $$\hat r=\cos\theta\hat i+\sin\theta\hat j,$$ and $$\hat \theta=-\sin\theta\hat i+\cos\theta\hat j.$$ Therefore $$\frac{d\hat r}{d\theta}=\hat\theta$$


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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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