Instantaneous velocity, on the other hand, describes the motion of a body at one particular moment in time. Acceleration is a vector which shows the direction and magnitude of changes in velocity. Its standard units are meters per second per second, or meters per second squared. (this is for number 3)
Given :
A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.
To Find :
The coefficient of static friction between the box and the plane.
Solution :
Vertical component of force :
![mg\ sin\ \theta = 120\times 10 \times sin\ 47^\circ{}=877.62 \ N](https://tex.z-dn.net/?f=mg%5C%20sin%5C%20%5Ctheta%20%3D%20%20120%5Ctimes%2010%20%5Ctimes%20sin%5C%2047%5E%5Ccirc%7B%7D%3D877.62%20%5C%20N)
Horizontal component of force(Normal reaction) :
![mg\ cos\ \theta = 120\times 10 \times cos\ 47^\circ{}=818.40 \ N](https://tex.z-dn.net/?f=mg%5C%20cos%5C%20%5Ctheta%20%3D%20%20120%5Ctimes%2010%20%5Ctimes%20cos%5C%2047%5E%5Ccirc%7B%7D%3D818.40%20%5C%20N)
Since, box is on the verge of slipping :
![mg\ sin\ \theta= \mu(mg \ cos\ \theta)\\\\\mu = tan \ \theta\\\\\mu = tan\ 47^o\\\\\mu = 1.07](https://tex.z-dn.net/?f=mg%5C%20sin%5C%20%5Ctheta%3D%20%5Cmu%28mg%20%5C%20cos%5C%20%5Ctheta%29%5C%5C%5C%5C%5Cmu%20%3D%20tan%20%5C%20%5Ctheta%5C%5C%5C%5C%5Cmu%20%3D%20tan%5C%2047%5Eo%5C%5C%5C%5C%5Cmu%20%3D%201.07)
Therefore, the coefficient of static friction between the box and the plane is 1.07.
Hence, this is the required solution.
Answer: As Earth spins on its axis, we, as Earth-bound observers, spin past this background of distant stars. As Earth spins, the stars appear to move across our night sky from east to west, for the same reason that our Sun appears to “rise” in the east and “set” in the west.
Explanation: