2008年11月19日 · If Olm's law is valid and R is constant then we know Voltage is a function of current so P lost (I)=V(I)*I. As you can see we have just expressed power lost completely as a function of current. Thus when Olm's law applies power loss is dependent on current. When we actually finish the substitution we get P lost =I 2 R.

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2015年6月21日 · There are several ways to write centripetal acceleration $$\\frac{v^2}{r} = \\omega^2 r = v \\omega$$ Are there intuitive explanations for any of these three forms? For instance, I can sort of expl...

an = rω^2 = r(v/r)^2 (replacing ω by the above equation) = v^2/r Share. Cite. Improve this answer ...

2010年11月17日 · Ac = V 2 /r and directly related to the radius in the equation: Ac = 4pi 2 r/T 2 is because in the first equation if you keep the velocity constant the bigger the radius the slower the object is going to move and the smaller the centripetal acceleration is going to be because that object is going to change its direction at a slower rate.

2020年12月3日 · $\frac{r}{C} = \frac{r}{2\pi r} = \frac{1}{2\pi}$ That 1 in the numerator is the one radian, and $2 \pi$ in the denominator is the what it takes to get around the circumference. So really, a single radian represents one radius of arc length if you define $2 …

2017年8月15日 · In the B field of the loop, if R is the radius and z is the distance along the axis perpendicular to the center of the loop, let z go to zero, and multiply by N loops. Starting with the B field of ...

2021年4月16日 · Using the equipartition theorem we have: $\langle E \rangle = N(\frac{3}{2})k_B T$ or $\langle E \rangle = \frac{3} {2}RT$. The factor of $3N$ arise from the number of molecules and the fact that we have three dimensional space, then we have three degrees of freedom from the momentum of each molecule.

2014年5月1日 · When to use P=I^2R, P=VI, P=V^2/R?? Hi, I know that P = I^2R is to find heat loss. but can we use P = V^2/R or P = IV to find the heat loss? simply when do we use which?? (or can we use all anytime?) The question said to find 1) the power GENERATED by a dc voltage source and to find the 2)...

2013年1月5日 · u=3GM^2/5r. The derivation follows from the volume of the sphere, 4 pi r^3 and area of the sphere 4 pi r^2. In Faber Jackson relation the gravitational potential of a mass distribution of radius R and mass M is given by the expression: U=-alphaGM^2/R where alpha=3/5 and the equation stands as: U=-3/5GM^2/R So I asked you the above question.