Car A is stopped at a traffic light. At the moment it starts to move again, it is situated 330 m from car B, which is traveling in the opposite direction at a constant velocity of 18.0 m/s. If car A accelerates at a rate of 3.0 m/s^2, where and after how long will the two cars pass each other?​
Please answer this question with steps..​

hopr

Explanation:

Let t be the time it takes for the car to catch up with the truck. Using kinematics equations, Distance by car = ut + 0.5at² = 0 + 0.5 * 3.72 * t². Distance by truck = 13.8 * t. Since these distances have to be equal, we have: 0.5 * 3.72 * t² = 13.8t, 1.86t² = 13.8t, t ≈ 7.42s. Hence, the distance is 13.8 * (7.42) ≈ 102.4m.

that would be b i just took the test : )

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards