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Ultimate shear stress of Mild steel is 210N/sq mm
Well, the typical bowling ball weight is between 8 - 16 lbs, and the length of a bowling lane is 60ft from the Foul line to the Pins. Let's do some math. FΔt = Δp F = Δp / Δt = [mv(f) - mv(i)] / t = [(7.0kg)(4.5m/s) - (7.0kg)(6.0m/s)] / 0.05s = -210N That means that a bowling ball is exerting about 210N of force. That's pretty self explanatory right there.
In the line of tropic of Capricorn with Atlantic ocean on the west,Angola in north,Botswana on east and south Africa in the south between approx 25 oE 11oE 210N 29oS.
Infinity Block didn't give enough dimension constrain. It could have one side a pin head with extreme length and place block on one side having pressure = Force/Area. If the Area is very small almost zero in area then we got infinity of pressure. If it is a cube, then it constrain the side and we only need to presume the density of metal. Probably try Plutonium or Uranium for highest density. Or may be it could define that block need to be made of at least 4 atoms to form the 4 corner and thus we define the minimum area a block can be. From this latest assumption, atomic radius of metal element presumably around 200 picometre. Distance from any corner is then = 4 x 200 = 800 picometre. The total area on one side is then 800 x 800 = 640000 square picometre In normal unit then it is 6.4 x 10-19 m2 The maximum pressure a block may exert when placed on one of it surface is then 210 N/ 6.4 x 10-19 m2 = 3.3 x 1020 N/m2 Compare to atmospheric pressure, it is 3,238,182,177,045,300 ATM as possibly the highest pressure it can exert.