AN IRON CUBE OF SIDE 10cm IS KEPT ON A HORIZONTAL TABLE.IF THE DENSITY OF IRON IS8000 kg / m3. FIND THE PRESSURE ON THE PORTION OF TABLE WHERE CUBE IS KEPT? (g - 10 m/s2)
PLEASE EXPLAIN BRIEFLY WITH STEPS
Side of cube = 10cm = 0.1m
Density = Mass / volume
8000 = mass / 0.1 * 0.1 *0.1
mass = 8000 * 0.001
= 8 kg
weight = mg
= 8*10 = 80N
Area of contact = 0.1 * 0.1 = 0.01
Pressure = 80 / 0.01
= 8000 Pa