uppose a 1,500-kg car passes over a bump in a roadway that follows the arc of a circle of radius 20.0 m as in the figure shown below.A car traveling to the right over a hump on the road at a velocity vector v.(a) What force does the road exert on the car as the car passes the highest point of the bump if the car travels at 8.90 m/s? (Neglect any friction that may occur.)magnitude Your response is off by a multiple of ten. kNdirection (b) What is the maximum speed the car can have without losing contact with the road as it passes this highest point? m/s

Questions 7 & 8 refer to the following in formation:A car of mass 760 kg travels along a flat road approaching a school crossing and hits a speed bump with a radius of 19.2 m.Question 7(3 marks)Calculate the maximum speed that the car can have in order to remain in contact with the speed hump.

A car goes along the circular curve of radius 30 m. If the road is banked at an angle of015 withthe horizontal, calculate the maximum speed in m/s at which the car can travel without skid.

Let's solve the homework question step by step. ### Part (a): Free Body Diagram and Kinetic Diagram at Point BAt point B, the car is at the top of the bump. The forces acting on the car are: 1. The gravitational force (weight) acting downward, \( F_g = mg \). 2. The normal force \( N \) exerted by the road on the car, acting upward. Since the car is about to leave the surface, the normal force \( N \) will be zero. The only force providing the centripetal acceleration is the gravitational force. ### Part (b): Instant Speed at Point BTo find the speed at point B such that the car is about to leave the surface, we use the centripetal force equation. At the top of the bump, the centripetal force is provided by the gravitational force: \[ F_g = \frac{mv^2}{r} \] Where: - \( F_g = mg \) is the gravitational force. - \( m \) is the mass of the car. - \( v \) is the speed of the car at point B. - \( r \) is the radius of curvature of the bump. Setting the gravitational force equal to the centripetal force: \[ mg = \frac{mv^2}{r} \] Solving for \( v \): \[ g = \frac{v^2}{r} \] \[ v^2 = gr \] \[ v = \sqrt{gr} \] Given: - \( g = 9.8 \, \text{m/s}^2 \) - \( r = 20 \, \text{m} \) \[ v = \sqrt{9.8 \times 20} \] \[ v = \sqrt{196} \] \[ v = 14 \, \text{m/s} \] So, the instant speed of the car at point B is \( 14 \, \text{m/s} \). ### Part (c): Constant Acceleration from A to BTo find the constant acceleration needed to change the speed from \( 30 \, \text{m/s} \) at point A to \( 14 \, \text{m/s} \) at point B over a distance of \( 100 \, \text{m} \), we use the kinematic equation: \[ v_f^2 = v_i^2 + 2a d \] Where: - \( v_f = 14 \, \text{m/s} \) is the final speed at point B. - \( v_i = 30 \, \text{m/s} \) is the initial speed at point A. - \( a \) is the constant acceleration. - \( d = 100 \, \text{m} \) is the distance. Rearranging to solve for \( a \): \[ a = \frac{v_f^2 - v_i^2}{2d} \] Substituting the given values: \[ a = \frac{14^2 - 30^2}{2 \times 100} \] \[ a = \frac{196 - 900}{200} \] \[ a = \frac{-704}{200} \] \[ a = -3.52 \, \text{m/s}^2 \] So, the car needs to decelerate at a constant rate of \( -3.52 \, \text{m/s}^2 \) to change its speed from \( 30 \, \text{m/s} \) to \( 14 \, \text{m/s} \) over a distance of \( 100 \, \text{m} \).

Question 8(2 marks)The car continues along the road, now travelling at 60.0 km h-1. There is another speed hump. As the car travels over this speed hump, the driver just starts to feel ‘weightless’. Calculate the radius of curvature of this second speed hump.

A car of mass 1400 kg drives in a circle of radius 50 m. If the car has a speed of 10 m/s, what is the centripetal force acting on the car?A.1400 NB.280 NC.2800 ND.140 N