A square solid iron block with a side length of 10 cm is placed in the center of the horizontal table top with an area of 0.2 m2 to calculate the mass of the iron block and the pressure on the table top The density of iron is 7.9 * 10 cubic

A square solid iron block with a side length of 10 cm is placed in the center of the horizontal table top with an area of 0.2 m2 to calculate the mass of the iron block and the pressure on the table top The density of iron is 7.9 * 10 cubic

L=10cm=1dm
m=pV=PL^3
p=7.9*10^3kg/m^3=7.9kg/dm^3
m=(7.9kg/dm^3)*(1dm)^3=7.9kg
Sectional area of iron block s = L ^ 2 = 1dm ^ 2 = 0.01M ^ 2

A square with a side length of 10 cm and a weight of 60 n is placed on a horizontal tabletop with an area of 0.6 square meters, and the pressure of the object on the tabletop is reduced
A. It must be 6000pa B. it could be 8000pa
Master answer, say the reason

B
P=F/S=60N/0.01㎡=6000Pa
I think you can do it. It's just a situation
Another case:
P=F/s s=F/P=60N/kg/8000Pa=0.0075㎡
At this time, the contact area between the cube and the desktop is 0.0075 m2, but the answer is limited by the "must" of option a, so choose "B". This is a special example

Put a 10 cm square iron block on each side on the center of a horizontal table with an area of 0.2 m2 to calculate the mass of the iron block

The volume of iron block is 0.001m3 (the third power of the edge), the density of iron is 7.8kg/m3, the mass of iron block = volume × density = 0.0078kg, which has no relationship with the tabletop area

A uniform cube with 100N weight and 20cm side length is placed in the middle of the horizontal square small desktop. If the side length of the small desktop is 10cm, the pressure of the cube on the desktop is zero______ The pressure is______ ．

The pressure of the cube on the desktop is equal to the gravity of the cube, f = g = 100N, the stress area is s = L2 = (0.1M) 2 = 0.01m2, the pressure of the cube on the desktop is p = FS = 100n0.01m2 = 10000pa