# Tag Info

1

No, not at all. Imagine doing the line integral of a vector field around a square path. Now envisage that the vector field varies in magnitude and/or direction with position. There is no particular reason that the line integral along one side of the square should cancel with that on the opposite side. In fact, the expression of Stokes's theorem that you ...

0

No it is not zero. It would have been zero had it been a conservative force e.g. gravity, electric field due to a conductor (electrostatic). But this field is induced electric field $\vec E$ produced due to changing magnetic flux and this field is non-conservative in nature. Hence like non-conservative forces, line integral around a closed loop is not zero ...

1

The integral of $sin(\theta)$ from 0 to $2\pi$ is zero. We can calculate it by $\int_{0}^{2\pi} sin(\theta)d\theta = [-cos(\theta)]\Big|^{2\pi}_{0} = -cos(2\pi) + cos(0) = -1 + 1 = 0$. So when you integrate to find $E_{r}$, you will find that it equals zero at the center of the circle. Also, you mentioned the integral from 0 to $\pi$ is 2. ...

0

$\left\langle \psi ,B\psi \right\rangle =\int{dr{{\psi }^{*}}i{{\partial }_{r}}\psi }=\left[ {{\left| \psi \right|}^{2}} \right]_{0}^{\infty }-\int{dri\left( {{\partial }_{r}}{{\psi }^{*}} \right)\psi }=\int{dr{{\left( i{{\partial }_{r}}\psi \right)}^{*}}\psi }=\left\langle B\psi ,\psi \right\rangle$ By definition, the hermitian conjugate satisfies: ...

2

The answer that you calculate is correct, but I think some clarification is in order. First, notice that you are asked to integrate a line segment, not a closed curve, so you should have no reason to expect that the answer should be zero. Second, $\oint \vec{F} \cdot ds=0$ only for conservative vector fields $\vec{F}$. Being a conservative vector field means ...

1

A line segment is not a closed curve, so your integral is not an integral over a closed curve, it is just a line integral; so your answer is correct. However, if you must insist that your integral is an integral over a closed curve, then you can consider the (degenerate) closed curve formed by going from A to B and then from B to A, in which case your result ...

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we first define velocity with the following equation $$v=\int{a}dt$$ Differentiating the equation would yield acceleration which is incorrect based on what the question is asking You stated your equation for velocity is velocity$$v=f(t)=2cm/s^3t^2+5cm/s$$ Therefor by simple substitution we just sub $v=f(4)$ $$f(t)=2\times(4)^2+5=37cm/s$$ Verifying the ...

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When you have a function $f(t)$ that expresses a quantity (say the velocity $v$), then you just evaluate that function at $t$ in order to get the function (velocity) at $t$. However, if I give you the position at $t$, $x(t)$, and I ask you for the velocity, then you first have to recall that velocity is the derivative of position with time: $$v = ... 1 Let's approach this slightly differently; I am sure you would agree that:$$v a = a v$$This is the equivalent of your equation after being multiplied by v where v=\frac{dx}{dt} and a=\frac{dv}{dt}. Now integrate this with respect to time:$$\int_{t_1}^{t_2} v a dt = \int_{t_1}^{t_2} a v dt  From the definition of the derivatives we know that ...

1

The answer by ryanp16 gives a great derivation of the equation you're asking about using calculus, and in fact, that is the approach that I would have taken had I not seen his answer. However, if you're not familiar with calculus, there's a second, algebraic approach that you can use to arrive at the same conclusion. I like to take this approach with high ...

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Acceleration $a$ is defined as the rate of change of velocity $v$ with respect to time $t$, or $a=\frac{dv}{dt}$. For constant acceleration we can integrate both sides with respect to $t$ to obtain $v=u+at$ where $u$ is the velocity at our initial time $t=0$. Since velocity $v$ is defined as the rate of change of displacement $s$ with respect to time we ...

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