Yahoo Answers: Answers and Comments for The speed of a particle moving in a circle 2.0 m in radius increases at the constant rate of 4.4 m/s2. At an i? [Physics]
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From Anonymous
enUS
Tue, 06 Jul 2010 20:20:34 +0000
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Yahoo Answers: Answers and Comments for The speed of a particle moving in a circle 2.0 m in radius increases at the constant rate of 4.4 m/s2. At an i? [Physics]
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From mcmannes: Newton's 1st regulation applies to moving ...
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Tue, 18 Oct 2016 04:20:00 +0000
Newton's 1st regulation applies to moving products  it says some component like "has an inclination to stay in action on a similar speed, in a at as quickly as line, till at last acted on by utilising skill of an outdoors rigidity". for this reason, on a similar time as an merchandise is going in a circle, it ought to prefer to have an outdoors rigidity appearing on it to reason the cost vector to maintain changing direction. That rigidity is termed centripetal rigidity on a similar time as the article is going in a circle. That rigidity motives acceleration it is termed centripetal acceleration. the linked cost of the centripetal acceleration, Ac, is given by utilising skill of Ac = m*w^2*r the placement m is the mass, w is the angular speed in radians/sec, and r is the radius. If Ac is shrink in 0.5 and the mass and radius stay consistent, w^2 ought to prefer to be shrink in 0.5. So if w_new^2 = (a million/2)*w_initial^2 w_new = 0.707*w_initial So the angular speed ought to prefer to be prolonged by utilising skill of a component of 0.707. The question is looking relating to the era and era is a million/frequency or a million/f. So shall we prefer to suitable prevalent the relationship between f and w. it fairly is two*pi*w = f So if angular speed ought to prefer to be prolonged by utilising skill of a component of 0.707, frequency ought to prefer to be prolonged by utilising skill of a component of 0.707. once you think approximately that era is a million/frequency or a million/f, the era ought to prefer to be divided by utilising skill of a component of 0.707, or prolonged by utilising skill of one million.414.

From anil bakshi: total acceleration = a = vector = a(centripeta...
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Tue, 06 Jul 2010 21:24:18 +0000
total acceleration = a = vector = a(centripetal) ( i) + a (radial) j = v^2/r ( i) + (dv/dt) (j)
i & j are unit vectors along radial and transverse sense
these 2 accelerations are at right angles, so by pythagorous
 a  = sqrt [{v^2/r }^2 + (dv/dt)^2 ] = 6
given dv/dt = 4.4
[{v^2/r }^2 + (4.4)^2 ] = 36
{v^2/r }^2 = 16.64
v^2/r = 4.08
v^2 = 4.08 * 2 = 8.16
v = speed of particle = 2.86 m/s