The base of a rectangular vessel measure 10m by 18cm. water is poured into a depth of 4cm. (a) what is the pressure on the base?. (b) what is the thrust on the base?.

a). The pressure that lies on the base would be:

b). The thrust that exists on the base would be:

a). Given that,

Length of the rectangular vessel

The Breadth of the rectangular vessel

Area of the vessel ×

×

To find the pressure,

Pressure(p) = Density × g × h

With

Density

Gravity(g) =

Height(h) = or

So,

P × ×

∵

b). Since,

Pressure

⇒ ×

⇒ ×

∵ Thrust or Force

Learn more about "Pressure" here:

link

a) P =392.4[Pa]; b) F = 706.32[N]

Explanation:

With the input data of the problem we can calculate the area of the tank base

L = length = 10[m]

W = width = 18[cm] = 0.18[m]

A = W * L = 0.18*10

A = 1.8[m^2]

a)

Pressure can be calculated by knowing the density of the water and the height of the water column within the tank which is equal to h:

P = density * g *h

where:

density = 1000[kg/m^3]

g = gravity = 9.81[m/s^2]

h = heigth = 4[cm] = 0.04[m]

P = 1000*9.81*0.04

P = 392.4[Pa]

The force can be easily calculated knowing the relationship between pressure and force:

P = F/A

F = P*A

F = 392.4*1.8

F = 706.32[N]

answer; how far out of place which the object finished relative to the starting location;