The gradient of a displacement-time graph gives
1 answer:
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F=1/(4*pi*e0)*q1*q2/r^2
e0 ≈ 8,854187817·10^−12
F=1/(4*pi* 8,854*10^(-12))*2*10^(-7)*2*10^(-7)/(0,3)^2 H = <span> 0,003995 </span>H ~ 0,004 H
e0 ≈ 8,854187817·10^−12
F=1/(4*pi* 8,854*10^(-12))*2*10^(-7)*2*10^(-7)/(0,3)^2 H = <span> 0,003995 </span>H ~ 0,004 H
Answer:
The angle of inclination of the rider to the vertical is 32.5⁰.
Explanation:
Given;
linear speed of the motor cycle, v = 90 km/hr = 25 m/s
radius of the circle, r = 100 m
The angle of inclination of the rider to the vertical is calculated using banking angle formula;
Therefore, the angle of inclination of the rider to the vertical is 32.5⁰