Question

weegy a 7.5kg block is placed on a table. if its bottom surface area is 0.6m2 , how much pressure does the block exert on the tabletop?

Answer 1

The pressure exerted by the block on the table is given by:

$p=\frac{W}{A}$

where W is the weight of the box, and A is the bottom surface area of the box.

The weight of the box is: $W=mg=(7.5 kg)(9.81 m/s^2)=73.6 N$

Substituting into the first equation, we find the pressure:

$p=\frac{W}{A}=\frac{73.6 N}{0.6 m^2}=122.7 Pa$