A car moving with a speed of 54 text{ km/h} t | vedclass

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(C) First,convert the speed of the car from $	ext{km/h}$ to $	ext{m/s}$:$v = 54 	imes \frac{5}{18} = 15 	ext{ m/s}$.Next,calculate the centripetal

Vedclass · Accepted Answer

$	an^{-1}(1.125)$

Vedclass · Answer

(C) First,convert the speed of the car from $	ext{km/h}$ to $	ext{m/s}$:$v = 54 	imes \frac{5}{18} = 15 	ext{ m/s}$.Next,calculate the centripetal acceleration $(a)$ experienced by the pendulum inside the car:$a = \frac{v^2}{R} = \frac{15^2}{20} = \frac{225}{20} = 11.25 	ext{ m/s}^2$.When the car turns,the pendulum experiences a pseudo-force in the horizontal direction,causing it to deflect by an angle $	heta$ from the vertical.The relationship between the angle $	heta$,centripetal acceleration $a$,and acceleration due to gravity $g$ is given by:$	an 	heta = \frac{a}{g}$.Substituting the values:$	an 	heta = \frac{11.25}{10} = 1.125$.Therefore,the angle $	heta = 	an^{-1}(1.125)$.

$A$ car moving with a speed of $54 \text{ km/h}$ takes a turn of radius $20 \text{ m}$. $A$ simple pendulum is suspended from the ceiling of the car. Determine the angle made by the string of the pendulum with the vertical during the turning. (Take $g = 10 \text{ m/s}^2$)

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