A car of mass 1000,kg negotiates a banked cur | vedclass

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(B) For a car moving on a banked road without friction,the condition for safe turning is given by the formula: $	an 	heta = \frac{v^2}{Rg}$.Here,the

Vedclass · Accepted Answer

$30$

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(B) For a car moving on a banked road without friction,the condition for safe turning is given by the formula: $	an 	heta = \frac{v^2}{Rg}$.Here,the banking angle $	heta = 45^\circ$,the radius of the curve $R = 90\,m$,and the acceleration due to gravity $g = 10\,m/s^2$.Substituting these values into the formula:$	an 45^\circ = \frac{v^2}{90 	imes 10}$.Since $	an 45^\circ = 1$,we have:$1 = \frac{v^2}{900}$.$v^2 = 900$.$v = \sqrt{900} = 30\,m/s$.Therefore,the maximum speed of the car is $30\,m/s$.

$A$ car of mass $1000\,kg$ negotiates a banked curve of radius $90\,m$ on a frictionless road. If the banking angle is $45^\circ$,the maximum speed of the car is ............ $m/s$ $[g = 10\,m/s^2]$.

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$A$ car of mass $1000 \, kg$ negotiates a banked curve of radius $90 \, m$ on a frictionless road. If the banking angle is $45^\circ$,the speed of the car is ......... $m \, s^{-1}$.

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