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2

You are correct, electric current consists of electrons travelling from one place to another. Some materials conduct electricity better than others. Copper is one of the best and that's why our conductors are usually made of copper. Aluminium is also very good (so is silver) and high-voltage cables are usually made of aluminium. However, everything conducts ...

0

Ground is at 0 potential,so,it accepts electron from negative terminal. And at very far place any positively charged electrode accepts electron from ground and current flow.

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Note that $\int^{\infty}_{-\infty}f(t)dt=\int^{\infty}_{-\infty}f(-t)dt$ and also $\int^{\infty}_{-\infty}\frac{df(t)}{dt}dt=\int^{\infty}_{-\infty}\frac{df(-t)}{-dt}dt$

1

Can an angle be defined as a vector? It depends on what you mean by "vector". If by "vector" you only mean something that has a magnitude and a direction, then yes, the axis-angle representation qualifies as a "vector". To a mathematician, a vector is something that is a member of a vector space. In this context, the axis-angle representation fails to ...

1

Describing a rotation as a vector, with the direction of the vector along the axis of rotation, and the magnitude of the vector as the angle, is known as the axis–angle representation.

0

This is a very interesting question which has at least two answers. The first answer is for an electrical/electronic engineer who wants to get a correct answer by using any systematic and coherent method. The Associated Variables Convention is used. You assign a current direction to each circuit element and then assign a plus sign at the point at which ...

2

The $dx$ is actually a vector $d\vec{x}$ (because $\vec{x}$ is actually a vector). A vector's sign depends on your coordinate axes, e.g. if you pick right to be positive then left pointing vectors are negative. In general, the work done $W$ is given by $$W = \int \vec{F} \cdot d\vec{x} = \int Fdx \cos(\theta)$$ where $\theta$ is the angle between ...

0

It all depends on the direction you assign while using $dx$. It is generally a vector, so the sign will automatically be adjusted according the operation used and its direction relative to other operand vectors. In your case, you are calculating the work done by the spring. The force due to a spring is always directed to its mean position. With this in ...

0

basic concept is to set up a coordinate system , Cartesian would be good here Here, I'm calculating work done BY the spring origin is the point where spring is in natural position (for us to use F = -Kx , otherwise, it would also contain some constant too, ex- if our spring's natural length is at (2,5) and spring only moves in along x axis direction, ...

0

You might get a wrong sign due to $cos(0)= 1$ and $cos(\pi)= -1$. You either take $x$ as a cartesian coordinate and integrate over it or you use polar coordinates with some maximum elongation $x$ which is constant. In your question you are mixing up both. The conventional sign here is $-$, however it depends on your way of counting the work, ie work done to ...

0

emf induced by inductor is negative but emf applied by battery on inductor is positive. So, that inductor tries net voltage across it is 0. Emf applied by battery is E°sinwt of which LdI/dt is applied across inductor, RI across resistance and Q/C across capacitor. -LdI/dt is induced across inductor and not applied. Now, consider direction of emf is ...

0

Sign conventions aren't "wrong", but they can be misleading. For example, we could re-define work done in a gravitational field so that escaping earth's gravity well would require negative work. That's an equally valid convention, but we associate positive work with effort, so reversing the convention would hinder our physical intuition for no discernible ...

2

The reason is mainly in order to be able to write the total energy of the system as $$E=T+V$$ where $T$ is the kinetic energy of the system. It is far more useful to choose the signs of the terms in the total energy to be $+1$ once and for all, rather than setting the force equal to plus the gradient of the potential. The link between the conventions is ...

4

So the BIPM has now released drafts for the mises en pratique of the new SI units, and it's rather more clear what the deal is. The drafts are in the New SI page at the BIPM, under the draft documents tab. These are drafts and they are liable to change until the new definitions are finalized at some point in 2018. At the present stage the mises en pratique ...

0

Energy or the value of $V(x)$ negative means it is a bound system. Think of it in this way, if a particle is free and has no kinetic energy and potential energy then it's total energy is zero. If this particle is not free or otherwise is bounded by a negative potential well then it's potential energy is $-V$. You have to give the same amount of energy, in ...

1

I think what is missing is the definitions of $F$ and $V$. Consider first the definition of potential energy. The potential energy at a point relative to another point is the work done by a external force (eg the force exerted by you on the mass, $\vec F_{my}$) in taking the mass from the first point to the second point. That force which you exert on the ...

3

I think your problem is in how you understand the gradient. Define your z-axis. Lets say it points up. Now $$V(z) = mgz$$. Thus gradient of V(z), that is $$\nabla V(z) = mg \hat z$$. Here, $$\hat z$$ is an unit vector pointing up. So the force will be a vector of length mg, pointing down, because of the minus sign.

0

This is happening because your field depends on $t$, $\Psi = \Psi (x,t)$. Therefore, when you perform the Wick rotation $t = i\tau$, you also Wick rotate to your field, and obtain an action for $\Psi(x,i\tau)$. In the second case, you obtain an action for $\Psi(x,-i\tau)$. Those are not the same $\Psi$'s.

-2

Answer you people have given is right, but it is from mathematical background. Physically, conservative force is a dissipative force. Due to dissipation properties we right as gradient of potential and negative sign comes from its opposite direction of action.

-1

I don't really remember the voltage laws, but I can tell you with only one source positive current always flows from plus to minus on a voltage source so Ia and Ib must be positive as youve drawn the arrows, and Ia is correctly negative relative to the "+" and "-" you've drawn on the 21kOhm resistor. I recommend not getting stuck up in the laws and ...

0

Kirchoff's laws say that the voltage on the bottom wire is some constant $V_0$, the voltage on the top-left wire is $V_0 + \mathsf{24~V},$ and Ohm's law gives us that the voltage on the top-right wire is $$V_0 + \mathsf{24~V} - 12 ~\mathsf{k\Omega}\cdot(I_a + I_b).$$ We can arbitrarily choose $V_0$ to be $\mathsf{0~V}$ or $\mathsf{-24~V}$ as you wish, but we ...

2

In statistical mechanics and field theory, the second type is referred to as a "connected" correlation function. You sometimes see the notation $$g_{\text{c}}(\mathbf{x}-\mathbf{x'},t-t') = \langle s_1(\mathbf{x},t) s_2(\mathbf{x}',t') \rangle_{\text{c}}\,\text{,}$$ where $\langle\ldots\rangle_{\text{c}}$ indicates that the product of the averages should ...

2

The second version is the first one but with the observables shifted by their respective averages: $$\langle \tilde A\tilde B\rangle = \langle (A-\langle A\rangle)(B-\langle B\rangle)\rangle = \langle AB\rangle - \langle A\rangle\langle B\rangle$$ Oftentimes in e.g. numerical studies, it is easier to just sample the average product of the original ...

2

The general Pearson correlation between two variables is defined as $$\textrm{cor}(X,Y) = E[XY] - E[X]E[Y]$$ up to a denominator containing the standard deviations of the distributions of the two variables. In some field theories the expectation values of the variable itself (one-point function) vanishes, therefore oftentimes the above definition reduces ...

3

The force in the middle spring is $k(y_2-y_1)$ because it is lengthened when $y_1\lt 0$ or $y_2 \gt 0$. A positive sign on that force indicates that $m_1$ is pulled down and $m_2$ is pulled up. When $m_2$ is pulled down, $y_2 \gt 0$ and there is a downward force on $m_1$, so it will cause an increase in $\ddot {y_1}$. It will also cause an upward force on ...

1

It seems to me that there should be a mistake. Let $A^{\mu}(x)=\partial^{\mu} \theta (x)$. Clearly you have $F^{\mu \nu}=0$. Thus, you just have to find a function $\theta(x)$ such that: (\partial_{\nu} \partial_{\mu} \theta)(\partial^{\nu} \partial^{\mu} \theta)\neq \frac{1}{2} (\partial_{\nu} \partial^{\nu} \theta)(\partial_{\mu} \partial^{\mu} \theta). ...

2

Horizontal position of indices matters in principle because one might want to raise and lower indices on the Christoffel symbols. If the horizontal position of indices are not observed in a consistent manner, it becomes ambiguous which index was raised or lowered, and so forth, in particular if the connection is not torsionfree. Also note that different ...

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