What pressure does a lead brick exert on the ground? PLEASE HELP!

A lead brick, 5.0 x12.0x19.0 cm, rests on the ground on its smallest face. What pressure does it exert on the ground? (Lead has a density of 11.8 g/cm3.)

This lead brick has a volume of 5*12*19 = 1140 cm^3. Given a density of 11.8 g/cm^3, the mass of the brick is 13452 grams, or 13.452 kg. The area of its smallest face is 5*12 = 60 cm^2. The mass (technically force due to weight, physics trolls) is distributed evenly over this area.

So, 13.452 kg / 60 cm^2 =

224.2 g/cm^2
21986.51 Pa
219.9 mB
3.189 psi
164.9 torr
0.217 atm

Pressure is Force divided by Contact area.

First we need mass, which is volume x density.
Volume = b x h x w = 1140cm^3
Mass = 1140 x 11.8 = 13452g or 13.452kg.
Force = m x a = 13.452 x 9.81 = 131.96N

It rests on its smallest face which is 0.05m x 0.12m = 0.006
Pressure = F/A = 131.96/0.006 = 21993.3N/m^2 or in sig figs 22000N/m^2.

N/m^2 can also be written as P for pascals.

Smallest face has an area of 5x12 = 60 cm² or 60e-4 m²
Volume is that x 19 or 1140 cm³
mass is 1140 cm³ x 11.8 g/cm³ = 13924 g or 13.9 kg
weight is 13.9 kg x 9.8 m/s² = 136 newtons
pressure in Pa or newtons/m² is P = 136 newtons / 60e-4 m² = 22700 Pa

edit, corrected math
.

Volume = 1,140cc.
Mass = (1,140 x 11.8) = 13.452kg.
Weight = (13.452 x 9.8) = 131.83N.

Area of small face in metres = (.05 x .12), = .006m^2.
Pressure exerted = (131.83/.006) = 21,971.6Pa, = 21.972 kPa.