An iron cube of side 10cm is kept on a horizontal table. If the density of iron is 8000kg/m^3. Find the pressure on the portion of the table where the cube is kept. (g=10m/s).
Answers
Answered by
174
Side of cube=1o cm=0.1m
Area of plane of cube=side x side=0.01m²
volume of cube=side³=0.001m³
given density=8000kg/m³
density=mass/volume
mass of cube= densityxvolume=8kg
force on the surface of table by iron block=mg=8x10=80N
Pressure=Force/area=80/0.01=8000N/m²
Area of plane of cube=side x side=0.01m²
volume of cube=side³=0.001m³
given density=8000kg/m³
density=mass/volume
mass of cube= densityxvolume=8kg
force on the surface of table by iron block=mg=8x10=80N
Pressure=Force/area=80/0.01=8000N/m²
Answered by
101
side = 10cm = 0.1m
volume = 0.1³ = 0.001 m³
density = 8000 kg/m³
mass = 8000×0.001 = 8 kg
weight = 8×10 = 80N
area of base = 0.1² = 0.01 m²
Pressure = W/A = 80/0.01 = 8000 Pa = 8kPa
volume = 0.1³ = 0.001 m³
density = 8000 kg/m³
mass = 8000×0.001 = 8 kg
weight = 8×10 = 80N
area of base = 0.1² = 0.01 m²
Pressure = W/A = 80/0.01 = 8000 Pa = 8kPa
Similar questions