The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.25 with the floor. If the train is initially moving at a speed of 48 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor

Explanation:

The forces acting on the crates when the train starts stopping are their weights, the normal force from the train, the static frictional force and the fictional force that is produced by the deceleration of the train. As the gravitational force, this fictional force is equal to the mass of the crates multiplied by the magnitude of the acceleration of the train. So, the equations of motion of the crates will be:

Since the static frictional force is , we get:

So we have a limit to the acceleration of the train. Now, we have to know the distance traveled by the train when it is stopping. Then, we use the kinematic formula:

Now we solve for the acceleration to combine this equation to the inequality we got before:

And solve for x:

Since we are looking for the minimum value for x, we consider the case in which that inequality becomes an equation:

Before we finish, we have to convert the unities of the initial velocity to meters per second:

Finally, we plug in the known values to get :

It means that the train can be stopped at a minimum distance of 36.2m at constant acceleration without causing the crates slide over the floor.

i dont know but

explanation:

whoever your teacher is, deserves a raise for using pokemon as an example

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.