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**r**is the center to center distance of the two masses. If the ...2 Answers · Science & Mathematics · 30/10/2014

At point

**r**the positive charge on the right creates a field kq/(**r**-a)^2, this field ...the right direction. The negative charge creates a field of magnitude kq/(**r**+a)^2, and this field is pointed to the left. To get the net field...5 Answers · Science & Mathematics · 31/01/2009

**R**= k.A.∆θ / L**R**depends on area A .. doubling dia. increases**R**by factor 4 .. (A proportional to D²)**R**depends on 1...1 Answers · Science & Mathematics · 17/11/2011

... the number of moles of moles in the gas, and so

**R**relates the energy of a mole of molecules to the temperature. ...4 Answers · Science & Mathematics · 13/01/2007

**R**is ion fact Avogadro's number multiplied by the...via Maxwell-Boltzmann statistics, if you must know, but essentially,**R**is slightly more fundamental than it initially appears.1 Answers · Science & Mathematics · 01/11/2013

...yet undetermined magnitude v and direction, draw an arc of a circle with radius of curvature

**r**and draw a vector V1 tangent to that arc. The back of the vector ...4 Answers · Science & Mathematics · 23/06/2011

...a long a circle would be given by <p> = <

**r***cos(Ө) ,**r***sin(Ө)> To get the velocity vector we take the...1 Answers · Science & Mathematics · 06/05/2012

vector addition

**r**= r1 + r2 resolve vector along (x) and (y) directions with respective unit vectors as (i) and (j)**r**(x) = r1 cos 59 (i) + r2 cos 55 (i)**r**(x) = 1.2 cos 59 (i) + 7.3 cos 55 (i)**r**(x) = [1.2 cos 59 + 7.3 cos 55] (i...1 Answers · Science & Mathematics · 04/07/2007

Multiply both sides by

**r**'(t). Then we have**r**'**r**'' = -MGr'/**r**^2. ... expression. The extra factor of 2 comes from integrating**r**'**r**'' after multiplying each side by**r**'. So yes, it...1 Answers · Science & Mathematics · 26/10/2014

My proof requires calculus. Consider any point (

**r**cos wt,**r**sin wt) on a circle with center (0,0) and radius**r**and rotational...2 wt] = sqrt[**r**^2 w^2] = rw^2, since cos^2 wt + sin^2 wt = 1. rw^2 =**r**(v/**r**)^2 = v^2 /**r**.3 Answers · Science & Mathematics · 02/04/2008