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Physics

Mechanics

Easy

Question

$T w o space p a r t i c l e s space o f space e q u a l space m a s s e s space a r e space r e v o l v i n g space i n space c i r c u l a r space p a t h s space o f space r a d i i$ $r equals fraction numerator 34.3 over denominator 2 pi end fraction$ and $v equals fraction numerator 2 pi r over denominator T end fraction equals fraction numerator 2 pi r over denominator square root of 22 end fraction$ $space space r e s p e c t i v e l y space w i t h space t h e space s a m e space s p e e d. space T h e space r a t i o space o f space t h e i r space c e n t r i p e t a l space f o r c e s space i s$

$θ = \tan^{- 1} (\frac{v^{2}}{r g}) = 45 °$
$M V^{2} / 4$
$M V^{2}$
$M V^{2} / 4$

The correct answer is: $theta equals tan to the power of negative 1 end exponent open parentheses fraction numerator v squared over denominator r g end fraction close parentheses equals 45 degree$

$T equals m omega squared r I f m a n d v a r e c o n s t a n t s t h e n space omega proportional to square root of T therefore omega subscript 2 over omega subscript 1 equals square root of 1 fourth end root not stretchy rightwards double arrow omega subscript 2 equals omega subscript 1 over 2 equals 5 r p m$

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The lens of a pair of spectacles used by a person is thinner in the middle and thicker towards the edges. The lens used in the spectacles is ____.

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