Problem Solving: Show your complete Solution.
2.A ball weighing 1.5 kg is whirled in a circular path at a speed of 12.0 m/s. If the radius of the circle is 5.6 m, what is the centripetal force?
1) a=v2/R=(2.2⋅106)2/(5.3⋅10−11)=9.13⋅1022 (m/s2)a=v^2/R=(2.2\cdot10^6)^2/(5.3\cdot10^{-11})=9.13\cdot10^{22} \ (m/s^2)a=v2/R=(2.2⋅106)2/(5.3⋅10−11)=9.13⋅1022 (m/s2)
F=ma=9.13⋅1022⋅9.11⋅10−31=83.17⋅10−9 (N)F=ma=9.13\cdot10^{22}\cdot9.11\cdot10^{-31}=83.17\cdot10^{-9}\ (N)F=ma=9.13⋅1022⋅9.11⋅10−31=83.17⋅10−9 (N)
2) F=mv2/R=1.5⋅122/5.6=38.57 (N)F=mv^2/R=1.5\cdot12^2/5.6=38.57\ (N)F=mv2/R=1.5⋅122/5.6=38.57 (N)
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