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16

Potential energy is still a scalar quantity even in more than one dimension. This is because it only has a magnitude, there is no direction of potential energy. You can think of it as similar to the temperature in a room. Even though it varies with position, it still does not have a direction.


10

No, just because a value changes over space doesn't mean it is a vector quantity. Potential energy is not a vector. It is a scalar quantity related to its corresponding conservative force by $$\mathbf F=-\nabla U=-\left(\frac{\partial U}{\partial x}\hat x+\frac{\partial U}{\partial y}\hat y+\frac{\partial U}{\partial z}\hat z\right)$$ So the force components ...


7

Since potential energy is function of position, ... hence can be considered as vector quantity? You make a conceptual mistake: "$n$-dimensional vector quantity" does not mean that a quantity depends on the position in an $n$-dimensional coordinate system. "$n$-dimensional vector quantity" means that a quantity requires $n$ different real ...


5

Area is sometimes a vector because a number alone does not include all of the information we need about those things called areas. The additional information is "which way is the area facing" - direction, which is exactly the kind of information that vectors convey very well. It may help to think about distance and displacement. In physics we ...


4

Definition: The change in potential energy of the system is defined as the negative of work done by the internal conservative forces of the system. Potential energy may vary with space just like mass of a non uniform rod which may be represented as $f(x,y,z)$. After all potential energy is basically negative of work done by internal conservative forces which ...


3

Area vector has a direction. It's direction is perpendicular to the surface. We know Force= Pressure*Area Now Force is a vector quantity , while pressure is a scalar quantity. So it's basically a constant K (pressure ) getting multiplied by area vector to give force vector. Note: it's not a dot product otherwise force would have been scalar.


3

Working in the (+ - - -) convention. For simplicity, assume the metric is diagonal (this proof still holds if it's not, it's just longer): \begin{equation} d \tau^2 = g_{00} dt^2 + g_{xx} dx^2 + g_{yy} dy^2 + g_{zz} dz^2 \\ 1 = g_{00} \left(\frac{dt}{d\tau} \right)^2 + g_{xx} \left( \frac{dx}{d\tau}\right)^2 + g_{yy} \left(\frac{dy}{d\tau}\right)^2 + g_{...


2

Current is a flow of charge, which can certainly have a direction. You could consider charge flowing left-to-right, or up-to-down, or any other direction in 3D space. Likewise areas can also be directed; they define a direction perpendicular to a surface. Current density is therefore a useful quantity that gives the current flowing through a surface, and the ...


2

There is no separate word to distinguish the vector of acceleration from its magnitude. The same is true with the word force, which is also both a vector and often described by the same word when talking of its magnitude. Velocity and speed seem to be the exception, probably because speed is an everyday term, " speed of going from town A to town B&...


2

The definition of proper time you used is wrong. To find the proper time, you can't just substitute it into the "regular" time. In special relativity, you don't notice this problem as $g_{00}$ has unitary norm, but in general relativity you have to remember that $ d\tau^2=ds'^2 = g_{\mu\nu}dx^\mu dx^\nu \Rightarrow u^{a} u_{a} =\frac{d x^{a} d x_{a}...


2

If a walker walks from $A$ to $B$, then from to $C$ to $D$ and then from $D$ to $E$, each walk can be represented by a vector, e.g. $$A \to B \Rightarrow \mathbf{AB}$$ Similarly we have the vectors: $$\mathbf{BC},\mathbf{CD},\mathbf{DE}$$ These can now be added up so that: $$\mathbf{AC}=\mathbf{AB}+\mathbf{BC}$$ $$\mathbf{CE}=\mathbf{CD}+\mathbf{DE}$$ ...


2

You may have been taught to skip a step. The cross product of two vectors does not equal the area of the parallelogram between them. It yields a vector whose magnitude is that area, and whose direction is defined to be normal to the surface. This magnitude is, of course, a scalar, as one would expect for an area. As others have mentioned, the direction of ...


2

Both equations are wrong. It looks like you drew free body diagrams and equated the opposing forces. This would only be valid when the weight of the pendulum is not accelerating. Let $\dot\theta$ denote the angular velocity, $\ddot\theta$ the angular acceleration and $R$ the length of the string. The centripetal acceleration along the tension force is $$a_{...


2

I did not follow your math, because frankly the algebra is tedious, but there are easier, less error-prone ways to derive the result you are after. Vector algebra identities are your friend. The helpful one here is the ubiquitous identity $$\vec A \times(\vec B \times \vec C) = (\vec A\cdot\vec C)\vec B-(\vec A\cdot\vec B)\vec C. $$ $$\vec\tau=I\oint{\vec r\...


2

Given that work is the area under the Force-Distance curve $$W = \int F(t)\,{\rm d}x$$ and power is the time derivative of work, and for each small-displacement increment ${\rm d}x = v\, {\rm d}t$ $$ P = \tfrac{\rm d}{{\rm d}t} W = \tfrac{\rm d}{{\rm d}t} \int F(t)\,v\,{\rm d}t = F(t)\,v $$


1

Will friction act in the south direction and the west direction respectively, or will friction act only in the south west direction combined and the block move in the north east direction with some acceleration ? As @Warrenmovic pointed out, the friction force will be in the opposite direction of the resultant force, i.e., $\sqrt 2$ N in the south west ...


1

This is how I deal with problem of finite friction. Assume friction is infinite and find the required friction force needed to resist all relative motion. Lets say this results into two co-planar friction forces $F_x$ and $F_y$. Calculate the magnitude of the required friction force $F = \sqrt{F_x^2+F_y^2}$ Compare the magnitude $F$ to the actual available ...


1

Friction always acts opposing the motion of the object. So if you have a resultant force in a particular direction, friction will always act in the opposite direction. So if your particle is moving north-west due to a force, friction will act to oppose that motion (hence south-east), hope that helps.


1

Current density $\bf J$ is related to current $I$ by $dI={\bf J\cdot dA}$. To get the current through a finite area, integrate $I$ over the area.


1

you have three points on a rigid body. $$\vec{R}_P\,,\vec{R}_A\,,\vec{R}_B$$ thus $$\vec{r}=\vec{R}_A-\vec{R}_P$$ $$\vec{r}'=\vec{R}_B-\vec{R}_P$$ and $$\vec{a}=\vec{r}-\vec{r}'=\vec{R}_A-\vec{R}_B$$ If only point P is changing with the time t; $\vec{R}_P=\vec{R}_p(t)$ thus $\vec{\dot{a}}=0$ , but if the body is rotate about axis that goes throw point P ...


1

The direction of instantaneous velocity at any time gives the direction of motion of a particle at that point in time. The magnitude of instantaneous velocity equals the instantaneous speed. This happens because, for an infinitesimally small time interval, the motion of a particle can be approximated to be uniform.


1

Yes. The tangential and normal components of the acceleration are orthogonal, so by the Pythagorean theorem, the magnitude of the resultant acceleration is $$\sqrt{(g \cos \theta)^2 + (g \sin \theta)^2} = g\sqrt{\cos^2 \theta + \sin^2\theta}=g.$$


1

Leonard Susskind's book is a very introductory book and therefore just half-justifies why the imaginary unit appears there. Actually it comes from a "workaround", the expressions of the $x$ and $y$ components of spin angular momentum is not direct. In the same fashion then, I'm gonna give my answer. You know that the probability of measuring spin ...


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