A 250 kg crate is placed on an adjustable inclined plane. If the crate slides down the incline with an acceleration of 0.7 m/s2 when the incline angle is 25ø, then what should the incline angle be for the crate to slide down the plane at constant speed?

21.2 degrees

Explanation:

Let gravitational acceleration constant g = 9.81 m/s2 directed downward. We can calculate the g component that is parallel to the 25 degree incline:

This parallel component would produce an acceleration of the block. But since the net acceleration is only 0.7 m/s2, there's a friction acceleration that hinders g parallel. This acceleration can be calculated by

The force of friction would be

Friction is a product of normal force and its coefficient

For the crate to slide down at constant speed, the net acceleration must be 0. In other words, the parallel g must be the same as acceleration caused by friction. Let this incline angle be α

answer: 0

explanation: im   14

5 w for 1hr = 5x60x60 joules=300x60=3000x6=18,000j as total energy conversion. efficiency of bulb needed for radiant to thermal within 18000