Answer
$(a) $ The direction is perpendicular to the road & in upward (Directed to sky)
$(b)\space19.9\space m/s$
$(c)\space 2.86\times10^{6}\space kgm^{2}/s$
Work Step by Step
Please see the attached image first.
(a) By using the right-hand rule we can get the direction - upward (Perpendicular to the road)
(b) Let's take,
Speed of the car = V
Angular momentum = L
Here we use the equation, Angular momentum (L) = mrV ; where, m - mass of the car, r - radius of circular turn, V - speed of the car
$L=mrV$ ; Let's plug known values into this equation.
$2.86\times10^{6}\space kgm^{2}/s=1150\space kg\times125\space m\times V$
$19.9\space m/s=V$
Speed of the car = 19.9 m/s
(c) We can write,
Angular momentum$ (L) = mrV$
$L=mRVcos\theta-(1)$
We can get,
$cos\theta=\frac{125}{R}=\gt R=\frac{125}{cos\theta}$
$(2)=\gt (1)$
$L=m\times\frac{125}{cos\theta}\times Vcos\theta=125mV$
This value is same to when the car moves on the circular turn.
Therefore,
Angular momentum $=2.86\times10^{6}\space kgm^{2}/s$
