New answers tagged

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So typically you would write $1.427 \pm 0.150 \cdot 10^3 \text{(units)}$ or so; technically you would put in parentheses but of course that’s up to you. Actually you would probably want to round to $1.43 \pm 0.15 \cdot 10^3 \text{(units)}$, since that difference of 0.02 standard deviations is not going to help anyone in being pedantic. Atomic/particle ...


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The term "uncertainty" is always vague, an umbrella term for cases that can be extremely different and therefore need to be approached differently. The first rule is always to clearly specify what kind of uncertainty we're speaking about. For example, it could be that you aren't able to measure some length $a$, but you know that's physically ...


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In case $x$ the system starts with Gravitational Potential Energy (GPE) of $-5$ and ends with GPE of $0$. So the GPE has been increased by $+5$ because $-5+5=0$. The amount of work done is $+5$. In case $y$ the system starts with GPE of $-50$ and ends with GPE of $0$. So the GPE has been increased by $+50$ because $-50+50=0$. The work done is $+50$. More ...


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To answer your question directly, the energy density is expressed in liters of battery. Lithium Ion batteries have a density of 250–693 W·h/L of battery.


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$\omega$ is a axial vector, not an ordinary or polar vector, and its sense depends on the handedness of the coordinate system. For a right handed coordinate system v=ω×r. See the textbook Symon Mechanics.


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In general, limits on an integral over a subset of $\Bbb R^n$ implicitly take care of integration direction. (The case $n=1$ is familiar; if $a<b$ the directions for $\int_a^bfdx,\,\int_b^afdx$ are obvious.) It doesn't matter whether what's integrated is a univariate dot product, a $3$-dimensional cross product or in general a $k$-dimensional integrand, ...


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The question implies that there is a mass at origin. Since gravitational force is always attractive and when we move from $\infty$ to $x$ we travel in the direction of the force, the cross product $\vec{E}.\vec{dx}$ is positive. Hence, $V(x)=-\int_{\infty}^{x}Edx$ The rest is according to the book. The question states that "...mass distribution is ...


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The final answer is written correctly, without a negative sign, as shown in the printed solution. The field is pointing in the +x direction. $\vec{E} = -\frac{\mathrm{d}U }{\mathrm{d} x} \hat{x}$ If the potential at x was negative as your are suspecting, and 0 at infinity, the derivative is > 0 and negative derivative is <0 , giving a field pointing ...


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While we derive the expression for line intergral we don't care for the sign of $\Delta x$ because that already taken into account with limits of integration.For example : consider the example given below in the comment box. So it's as given in the book.


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Let f be the magnitude of the force exerted by the particle at the origin on the particle being moved. If the force exerted by the particle at the origin on the particle being moved is attractive, then you need to exert a force in the positive radial direction to move it +dr, and the work you expend is +fdr (and the work the particle does on you is -fdr). If ...


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There is nothing wrong! The force is radially outward. If the particle travels from some point to infinite, then the work turns out to be positive. If the particle travel from infinite to some finite point the work is done by force will be negative. This can be understood from direction of force and displacement as you did. You can equally understand it as ...


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Other answers are good, but I think there is a misunderstanding of the OP that needs to be addressed. There is often an ambiguity on the meaning of $g$, that can be used for two related things: The scalar $g$, that it's about $9.81 \text{m}/\text{s}^2$ The vector $\vec{g}$, which points downward. $\vec{g}=(0, 0, -g)$ Please notice that in the convention ...


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Your question addresses the direction of gravitational force on the pendulum, but there is a deeper idea that defines the direction of gravity. The concept of the gravitational force that mass 1 exerts on mass 2 is inherently the concept of a force vector, made explicit by a mathematical formulation that references a unit vector pointing from mass 2 to mass ...


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I hope you are familiar with Cartesian Coordinate System According to this system, we take any vector directed downward and leftward to be negative, and any vector directed upward or rightward to be positive. The signs represent the direction of the vector(*). Since gravity always acts downward, towards the center of the Earth, we take it to be negative. ...


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As a matter of fact, we all know that gravity is downward (or more accurately towards the center of the earth). $$\mathbf{g}=-\frac{GM_e}{R_e^2}\hat{r}$$ if we negate that vector, then it should move the ball up. If there were no gravity (or that means you are in space), then there will no force, and the pendulum will remain in its initial position. If you ...


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This is something I'm personally also trying to understand better, but here is something that might help. I was able to derive the minus sign by starting with the bivector and then deriving the usual $\vec{B}$ pseudovector field from that. Since I start by assuming the bivector is positive, I get the result that the pseudovector points in the "opposite ...


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In reality, the movement of negative electrons is the true current, but as said in earlier answers by convention we model current flow (conventional current) as that of positive charges. Also see answers to Conventional current, electron flow and ammeter.


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Since up to now nowbody posted an answer I would like present a heurstical hand wavy argument which came to me into mind which migth be exactly that one which my lecturer also intended to use. I would be thankful if somebody could look through it and tell me if what I'm write now make any sense. Recall I asked why when we have our quantized KG-field $$\hat{\...


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$$\mathbf{F}\cdot d\mathbf{s}=|\mathbf{F}||d\mathbf{s}|\cos\theta$$ if $ds<0$, then $|d\mathbf{s}|=-ds$ ($d\mathbf{s}$ is a vector, notice the bolded $s$). We suppose that $x$ will vary from $8$ to $0$, obviously $dx<0$ (because $x$ is decreasing). $$\begin{align*} \int_{\mathbf{s}_i}^{\mathbf{s}_f}\mathbf{F}\cdot d\mathbf{s}&=\int_{\mathbf{s}_i}^{\...


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Your error is because your limits of integration don't agree with the sign convention for $\mathrm{d}x$ in 2C. When you do the work line integral from point '8' to point '0', the direction of $\mathrm{d}\mathbf{s}$ is in the direction $8\to0$, and is in the $-\mathbf{x}$ direction, which is the opposite direction as $\mathbf{F}$; hence, your equation should ...


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In the old days a variable resistor (or potentiometer or rheostat) was typically made using a coil of wire with a sliding contact that can be moved along the wire. There is a nice picture of this on the Components 101 site: The symbol for a variable resistor was derived from this: Life being too short the symbol was subsequently simplified to the form you ...


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Yes, you are correct. The "real" derivative has a downstairs index because that's just the way derivatives transform, so we always have $$\partial_\mu = \left(\frac{1}{c} \partial_t, \nabla\right)$$ and $\partial^\mu = \eta^{\mu\nu} \partial_\nu$, because that's what index raising means. So with some simple matrix multiplication we see that in the $...


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