A passenger in a car taking a turn at a speed of 10m/s pushes against the door with a 200-N force If the door opens under a force of 800 N. calculate the minimum speed for which the door will open throwing the passenger out of the car You can ignore the friction force between the passenger and the seat

To find the minimum speed at which the door will open and throw the passenger out of the car, we need to use the concept of centripetal force. The centripetal force is given by the formula: where:- is the centripetal force,- is the mass of the object (in this case, the passenger),- is the velocity of the object,- is the radius of the circular path.Given:- The passenger is pushing against the door with a force of 200 N.- The door opens under a force of 800 N.We need to find the minimum speed for which the door will open. Let's denote the mass of the passenger as and the radius of the circular path as .The centripetal force required to keep the passenger moving in a circular path is provided by the normal force exerted by the seat. This force can be calculated using the given forces: where:- is the normal force exerted by the seat,- is the force exerted by the passenger to push the door.Since the door opens under a force of 800 N, the normal force is 800 N.Therefore, the centripetal force is: Now, we can use the centripetal force formula to find the velocity : Solving for : To find the exact value of , we need the values of and . However, since the problem does not provide these values, we can't calculate the exact speed. But the formula above gives us the relationship between the speed and the given forces.