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1

The difference is that $dW$ is an infinitesimal ''quantity'', whilst $W$ is not. I assume the context here is thermodynamics, which make use of calculus. In calculus there is the concept of the infinitesimal. I suggest, for you, to concern yourself with the structure of calculus if you are to tackle thermodynamics.


0

Light years help give an idea of distances through space and time. When we look at a star 100 light years away, that 100 light years not only gives an idea of the immense distance to the object but will also tell us that what we see is light from 100 years in the past.


0

Ultimately, the answer boils down to convenience. When we want to describe the distance between here and, for instance, the star Sirius (the brightest star in our night sky), it would be a little cumbersome to write $\ell=8.13\times10^{18}\,{\rm cm}$ any time we want to write its distance from us. And really this goes for any astronomical object: they're ...


1

Whenever you see things at a distance, your perceptions are of things that happened in the past. You can see it at a football game, for example, where someone kicking the ball is seen very much before it is heard (sound is slower than light). Light only has a finite speed, too; and light turns out to be the fastest thing there is. A light year is the ...


0

You can't state what the potential energy of a system is in absolute terms, you can only talk about the difference in potential energy between two states. In your example, you are quite free to define the potential energy of two nearby particles having the same charge as anything you like, positive, negative or zero. For example, what is the potential ...


0

How do someone determine the direction of voltage arrows in a circuit? Is there a physical reason to choose a direction or the other? I always use just one simple rule in all talk and drawings about electricity and circuits: Arrows point in the direction a positive charge would move. For current arrows, this explains the direction from a positive to ...


1

In this case it's because of who's doing the work. The derivation you cite regards the work as being done by the gravitational field. This means in bringing the object from infinity, the gravitational field has lost that energy to the object. Conversely, the work done on the object is the positive value. Everything else works out and when we take the ...


3

She asked us if the body was accelerating or slowing down Acceleration is defined as the time rate of change of velocity and, in this example, the acceleration is constant and positive. So, the full answer is: the velocity of the body is always increasing while the speed is decreasing for $t<1$ and increasing for $t>1$. In this plot, the ...


0

I understand what your teacher is saying, but I think she's wrong. In my physics classes we were told never to use the word "de"celeration, only acceleration. Why? For elegance reasons mostly. Acceleration is a vector quantity, therefore its magnitude is always positive or zero with a direction. The sign (or direction) of the acceleration depends on your ...


3

The initial velocity and acceleration here are in opposite directions. The magnitude of velocity (represented by $S=|\vec v|$) decreases upto a certain instant. (i.e. where $\vec v=0$). Edit: Also, consider these graphs. ($t^.$ being the time where $v=0$) Note how the velocity increases but the magnitude of it (in the $S$, $t$ graph) decreases till $t^.$. ...


1

For a newtonian fluid, you can write the total stress tensor $\sigma_{ij}$ as $$\sigma_{ij} = -p \delta_{ij} + \tau_{ij}$$ with $$\tau_{ij} = \mu \left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right).$$ $-p \delta_{ij}$ is the classic pressure term and $\tau_{ij}$ is the shear stress tensor with $\mu$ the shear viscosity. When ...


0

As the weight hanging off the table falls, it has some acceleration downwards. The force on the body is entirely due to gravity and is known. The string acts as a linkage between the hanging mass and the mass on the table. Really in this case, we have weights $m_1$ and $m_2$ being pulled by a force $m_1g$, which leads to the acceleration of the system being ...


1

$U$ in your equation in potential energy, and $W$ is internal work. That is, the work done by forces within the system. The system in question here comprises the object and the earth, and the internal force is gravity. The work that you do to lift an object is external to the system, and does not appear in your formula. In your scenario, lifting an ...


6

Given the way that you've presented your table, I would personally put a "-" rather than a 1 in the units column. This to me would signify that units such as "g, km, s, A" etc. do not apply here. In terms of your symbols, in many branches of physics it is common to use a "hat", "tilde" or "star" notation above a symbol to indicate that it is a ...


5

Conventionally we use $1$ for dimensionless quantities, although it may cause some confusions. In additon, The International Committee for Weights and Measures contemplated defining the unit of 1 as the 'uno', but the idea was dropped. --https://en.wikipedia.org/wiki/Dimensionless_quantity


2

The only convention I am familiar with is to put a $1$ for dimensionless quantities. If you feel that this could give rise to confusion, you could explain the convention somewhere above or below the table.


3

Yes. In the unit column, put $1$.


1

It would help me, but even more importantly you, if you defined what your symbols are supposed to mean. You did that with $U_i$ and $U_j$. So let's be precise together, as an exercise: Let's explicitly state that we consider a point mass $m$ in a gravity field caused by a much larger mass $M_E$ at the origin of the coordinate system, so that we can assume ...


6

First, there is nothing wrong with our charge polarity conventions. They are predict electrical phenomena just as accurately as the opposite convention would have done. Did we have a problem if since the begin of their discovery we called them positive particles and negative to protons? We could predict the behavior of electrical phenomena equally ...


2

Let's drop the scaling constants and say that the wavefunction in momentum co-ordinates $\psi_p(\vec{p})$ is the FT of that $\psi_x(\vec{x})$ in position co-ordinates , i.e. $$\psi_p(\vec{p}) = \mathfrak{F}_{\vec{p}}(\psi_x(\vec{x}))$$ where $\mathfrak{F}$ is the Fourier transform. Now you propose to take the inverse Fourier transform of $\psi_x$. The ...


1

A is wrong because the concept of "at rest" implies an arbitrary decision. The right answer is B which is correct whatever the observer's chosen frame.


3

You are right in assuming that being at rest is considered a special case of rectilinear uniform motion (abbreviated r.u.m. hereafter). The closest thing to a reason for this convention I can give is this: It is a nice property to have all observers in inertial frames agree on whether a body is in r.u.m. However, for every such body in r.u.m. (which we ...


-1

A body with no forces on it always follows a geodesic. In a gravitational field, two nearby object start at rest with respect to the each other. One is above the other. The separation between them increases. Neither A nor B is correct. Edit Thank you pfnuesel. If general relativity is not considered, then B is correct. An object that starts at rest ...



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