Consider a sphere of radius r.
Let the z-axis be the vertical axis and suppose the sphere's centre is at z = 0. [Thus the sphere goes from z = -r to z = r.]
Suppose the volume of liquid in it is V.
Then you need to sove
3V/pi = 3r2z - z3 + 3r3 - r3 = 3r2z - z3 + 2r3 for z.
Then, the depth is z + r
volume = width x height x depth
length times width times depth
Three dimensional objects would be a sphere, cube, pyramid, cylinder, cones, etc. Anything with a width, height, and depth is three-dimensional.
That's the amount of water in a round pool whose radius is 'r' .
No, depth is not a vector.
v=2d/t
volume = width x height x depth
length times width times depth
L x B x D or Length times Bredth times Depth
Is the land flat? if so you'd simply multiply the area of land by the depth of sand.
A sphere is three-dimensional. It has height, width and depth.
Liquid pressure depends on depth. It can be calculated from liquid density times depth.
The length times width times the depth of the holes of the bench times the amount of emulsion in the holes
pressure of liquid on bottom=density*gravitational force*depth :)
Density of liquid (kg/m3) * gravitational constant (m/s2) * depth (m) = (extra) pressure under liquid (Pa) Density of water = 998 (kg/m3) gravitational constant = 9.81 (m/s2) 1,000 ATM = 1.013*10-5 Pa Raw formula: pressure under water = 1 (ATM) per 10 (m) depth
Average depth. Depth of the deep end @ 8'. Depth of the shallow end 3'. Add the two together and divide by 2 = average. Ken
The formula relating the pressure in a liquid to the depth of the liquid is P = P0 + dgh. P is the pressure, P0 is atmospheric pressure, d is the density of the fluid, g is the acceleration of gravity, and h is height below the surface of the water.