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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?.​

Respuesta :

rafaleo84

Answer:

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]

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

[tex]P =392.4[Pa][/tex]

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

[tex]F = 706.32[N][/tex]

a). Given that,

Length of the rectangular vessel [tex]= 10m[/tex]

The Breadth of the rectangular vessel [tex]= 18cm or 0.18m[/tex]

Area of the vessel [tex]= L[/tex] × [tex]B[/tex]

[tex]= 10[/tex] × [tex]0.18[/tex]

[tex]= 1.8m^2[/tex]

To find the pressure,

Pressure(p) = Density × g × h

With

Density [tex]= 1000 kg/m^3[/tex]

Gravity(g) = [tex]9.81m/s^2[/tex]

Height(h) = [tex]4 cm[/tex] or [tex]0.04m[/tex]

So,

P [tex]= 1000[/tex] × [tex]9.81[/tex] × [tex]0.04[/tex]

∵ [tex]P = 392.4[/tex]

b). Since,

Pressure [tex]= F/A[/tex]

⇒ [tex]F = P[/tex] × [tex]A[/tex]

⇒ [tex]F = 392.4[/tex] × [tex]1.8[/tex]

∵ Thrust or Force [tex]= 706.32 N[/tex]

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