A tall, open container is full of glycerine. At what depth h below the surface of the glycerine is the pressure 2170 Pa greater than atmospheric pressure? The density of glycerine is 1.26 × 10 3 kg/m 3 .

0.176 m

Explanation:

Given

Pressure of glycerine, p = 2170 Pa

Density of glycerine, ρ = 1.26*10^3 kg/m³

Depth, h = ?

To solve, we use the formula of pressure

P = ρgh

Where

ρ = density of glycerine

P = pressure of glycerine

g = acceleration due to gravity

h = depth of glycerine

2170 = 1.26*10^3 * 9.8 * h

2170 = 12348 * h

h = 2170 / 12348

h = 0.176 m

Therefore, the depth below the surface at which the pressure of glycerine is greater than the atmospheric pressure is 0.176 m

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a ball spinning on a finger   >   kinetic energy > moving energy

a ball rolling on the ground   >   kinetic energy   > moving energy

a   ball flying through the air   > kinetic energy   >   moving energy

hope that                                         : )

multiply 2kg by latent heat of vap