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 :

Horizontal component of force(Normal reaction) :

Since, box is on the verge of slipping :

Therefore, the coefficient of static friction between the box and the plane is 1.07.
Hence, this is the required solution.
Answer:
Explanation: When a body covers unequal distances in equal intervals of time in a specified direction, the body is said to be moving with a variable velocity. Example: A rotating fan at a constant speed has variable velocity, because of continuous change in direction.
Answer:

Explanation:
The rotated angle is given by:

Since this is a quadratic equation it can be solved using:

Rewriting our equation:


Since
we discard the negative solution.

Answer:
15.7m/s
Explanation:
To solve this problem, we use the right motion equation.
Here, we have been given the height through which the ball drops;
Height of drop = 14.5m - 1.9m = 12.6m
The right motion equation is;
V² = U² + 2gh
V is the final velocity
U is the initial velocity = 0
g is the acceleration due to gravity = 9.8m/s²
h is the height
Now insert the parameters and solve;
V² = 0² + 2 x 9.8 x 12.6
V² = 246.96
V = √246.96 = 15.7m/s