Question

please help me for this

could not getting it​

Answer 1

Solution :-

Provided :-

Mass = $10^{-10} \: kg$

Angular speed = 400 rad/s

Radius of circle = 0.2m

Now we have to find out the centripetal force.

$F_c = m\omega^2(r)$

Where

m = mass

ω = angular speed

r = radius of circle.

Now centripetal force

$F_c = 10^{-10} \times (400)^2 \times 0.2$

$F_c = 10^{-10} \times 160000 \times 0.2$

$F_c = 10^{-10} \times 1.6 \times 10^{5} \times 0.2$

$F_c = 1.6 \times 10^{-5} \times 2 \times 10^{-1}$

$F_c = 3.2 \times 10^{-6} \: N$

Answer 2

$\huge{\sf{\underline{Given:-}}}$

${ \tt{1.mass \: of \: particle(m) = 10 {}^{ - 10} kg}} \\ { \tt{2.angular \: speed( { \omega})\: = 400 \: rads {}^{ - 1} }} \\ { \tt{3.radius \: of \: circle(r) = 0.2 \:m}}$

$\huge{\sf{\underline{To\:Find:-}}}$

${\tt{Centripetal\:Force\:(F_{c}):-}}$

${ \tt{m{ \omega {}^{2} }r}}$

$\huge{\sf{\underline{Solution:-}}}$

${ \tt{10 {}^{ - 10} \times400 {}^{2} \times 0.2 }} \\ 10 {}^{ - 10} { \tt{\times 160000 \times 0.2}} \\ 10 {}^{ - 10} { \tt{\times 1.6 \times 10 {}^{5} \times 0.2}} \\ { \boxed{ \bf3.2 \times 10 {}^{ - 6}}}$

${\tt{Hence\:centripetal\:force\:is\: 3.2× 10^{-6}\:N}}$