Determine the pressure at a depth of 0.6 m in water, kerosene, mercury.

Water, kerosene and mercury are substances with different densities. The pressure directly depends on the density of the liquid and on the depth to which the investigated object is immersed.

So, the formula for determining pressure at depth is as follows:

P = ρ × g × h,

where: P – pressure;

ρ is the density;

g is the acceleration of gravity;

h – depth.

According to the Table of Density of Substances, we determine that:

– the density of water is 1000 kg / m3;

– the density of mercury is 13600 kg / m3;

– the density of alcohol, ethanol, kerosene is 800 kg / m3.

The acceleration due to gravity (g) is constant and equal to 9.8 N / kg.

The depth (h) is given to us in the problem statement and is equal to 0.6 m.

Having all the data, let’s substitute them into the formula:

P water = 1000 kg / m3 × 9.8 N / kg × 0.6 m = 5880 Pa.

P kerosene = 800 kg / m3 × 9.8 N / kg × 0.6 m = 4704 Pa.

P mercury = 13600 kg / m3 × 9.8 N / kg × 0.6 m = 79968 Pa.

Answer: The pressure at a depth of 0.6 m in water will be 5880 Pa; in kerosene – 4704 Pa; in mercury – 79,968 Pa.