Homework Statement: There are the propagation delays from High to Low and from Low to High, and I wonder if they are equal?
Homework Equations: That is
$$ t_{PLH}=t_{PHL}$$
I suppose they should be equal.
There is a property to geometric distribution, $$\text{Geometric distribution } Pr(x=n+k|x>n)=P(k)$$.
I understand it in such a way: ##X## is independent, that's to say after there are ##(n+k-1)## successive failures, ##k## additional trials performed afterward won't be impacted, so these ##k##...
You are totally right. I made a mistake that the inequality mentioned above cannot express the relationship between ##x## and ##y##, which is a significant constraint of ##y##.
Starting from the simple case, there is a single wave ##e=a\cos(2\pi ft+\frac{2\pi}{\lambda}x+\phi_0)##, and integrate in such a way, where ##T_{eye}## stands for the response time of human eyes' response time towards energy change:
$$I=\int_{0}^{T_{eye}}e^2dt$$
The calculation includes...
Thank you for your answer. I do observe my answers are not accurate; however, what is wrong with my method, or if there is any modification to make it right?
1. I consider this problem algebraically, ##c\cdot \vec{u}+(1-c)\cdot \vec{v}=c(1,2)+(1-c)(2,1)=(c,2c)+(2-2c,1-c)=(2-c,1+c)##; since the constraint I know is ##c\geq 0##, I can conclude the expected vectors##(x,y)## must have ##x\leq2, y\geq 1##.
2. Similarly, I get...
The additional law with two elements can be expressed $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$, while the law with three elements can be $$P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)$$
Now I wonder if there is the more general form of addition law, which applies...
Above is an example figure.
2. When a ring in a changing magnetic field is not complete (i.e. open circuit with a small gap), how to analyze the emf of the ring?
According to the general form of Faraday's law, ## \oint \vec{E} \cdot d \vec{s} = -\frac{d \Phi}{dt} ##, I deduce that although it...
The result is $$ v(t)=\frac{\varepsilon}{Bd} (1-e^{-\frac{B^2d}{mR}t}) $$
According to L'Hospital's Rule, ## \lim_{B \to 0} v(t)=0,## that is to say when B is sufficiently small, the bar won't move at all, and the whole circuit remains still. It is reasonable.
##l## is the same as ##d##.
Not solving for v directly was because I consider that ## v(t)=v_0-\int a(t)dt,## which contains ##\int vdt=x##; therefore, I wanted to solve the ODE for x first, which is shown above.
Fortunately, thanks to your hint, I manage to solve v directly based on Newton's...
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Above is the figure of the problem.
I am trying to solve x(t) and differentiate it to obtain v(t); however, I have difficulty solving the differential equation shown below.
$$ v(t)=\int a(t)dt=\int \frac{B(\varepsilon-Blv)d}{Rm}dt \Rightarrow \frac{dx}{dt}=\frac{B\varepsilon...
Such treatment simplifies (a) a lot, such that ## I=\frac{\varepsilon}{R_{60}}=1.43A.##
Meanwhile, in (b), ##\varepsilon_L=-\varepsilon_{R60}##, where ##\varepsilon_{R60}## represents the potential drop of the resistor of 60Ω, only when satisfying this condition, can the 15mH inductor be treated...