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6

On spherical coordinates, the gradient of a general function $V$ is: $$ \nabla V = \frac{\partial V}{\partial r}\mathbf e_r + \frac{1}{r}\frac{\partial V}{\partial\theta}\mathbf e_\theta + \frac{1}{r\sin\theta}\frac{\partial V}{\partial\phi}\mathbf e_\phi $$ If $V(r, \theta, \phi)$ only depends on $r$, that is $V = V(r)$, which is exactly the case of the ...


4

Given your question, it seems likely that your misunderstanding comes from a limited sense of vectors, fields, and partial derivatives. So there's a lot of education that we have to cover in a very short time. Multivariate functions When we transition from a function $f(x)$ to a field, which is a function of many variables $f(x, y, z)$, we suddenly have ...


2

If you really had a "constant power" engine, and all that power was transferred to your rocket which does not lose mass, it would result in a linear increase in the kinetic energy. And since the kinetic energy $E=\frac12 m v^2$, you can find the velocity at a given time from $$P\cdot t = \frac12 m v^2\\ v = \sqrt{\frac{2 \cdot P \cdot t}{m}}$$ If you ...


2

In those notes, it states that the left part of the string pulls the dot with a force proportional to the slope. However, the right side pulls the dot in the other direction with a force proportional to the slope, so that if the slope is constant, there is no net force on the dot. The only way to get a net force is for the slopes to be different on the left ...


2

In the systems you describe, each string connects always two masses. The tension force exerted on these two masses by the string is equal in magnitude and opposite in direction. Hence, the work done by each string on the two masses attached have opposite sign. As a consequence, if you sum up all contributions from each string and each mass, the net work ...


1

As said in the comments, there are only 4 known fundamental forces, gravity, strong and weak interaction, and electromagnetism. Since we are restricted to low energy classical mechanics, we can ignore two of them, and we are left only with gravity and electromagnetic forces (which includes, among others, magnetostatic and electrostatic forces). Moreover, ...


1

There is less recoil, and less chamber pressure. Repeating actions usually do not have enough energy to cycle. The noise is comparable to a live round. Note that a blank round does not mean "no discharge". More than a few people have been killed by the wadding and other debris - it just doesn't have any appreciable range.


1

As a bus passes you at speed, it displaces the air that occupied the volume which the bus fills. The displaced air flows past the bus, creating a high pressure area near the bus which pushes you away. After the midpoint of the bus has passed you, the displaced air starts to flow into the area vacated by the bus as it leaves. This creates an area of low ...


1

Let $F$ be a force field. Assuming that the force field is a conservative vector field, then it follows that the line integral of the force field is zero $$\oint_{O} F \cdot dr = 0$$ The del operator $\nabla$ is defined in 3 dimensions as $$\nabla = \langle\frac{\partial}{\partial{x}}, \frac{\partial}{\partial{y}}, \frac{\partial}{\partial{z}}\rangle$$ ...


1

The work-energy theorem leads us to the following result; \begin{equation} \oint \vec F\cdot d\vec s=0 \end{equation} \begin{equation} \oint \vec F\cdot d\vec s=\underbrace{\int \int }_{\text{surface}}(\nabla \times \vec F)\cdot d\vec n \end{equation} Using the rules of vector calculus there must exist some scalar function such that; \begin{equation} \vec ...



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