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A block of metal of density 3000 kg/m3 is 2m high and stands on a square base of side 0.5 m. Calculate:
1.The area of the base of the block
2.The volume of the block
3.Mass of the block
4.Weight of the block
5.The pressure exerted by the block

Respuesta :

Answer:

1)  A = 0.25 m², 2) V = 0.5 m³, 3)   m = 1500 kg, 4) W = 14700 N,

5)  P = 58800 Pa

Explanation:

1) The area of ​​the base is square

          A = L²

         A = 0.5²

         A = 0.25 m²

2) The block is a parallelepiped

         V = A h

         V = 0.25 2

          V = 0.5 m³

3) Density is defined

           rho = m / V

           m = rho V

           m = 3000 0.5

           m = 1500 kg

4) The weight of a body is

           W = mg

            W = 1500 9.8

            W = 14700 N

5) The pressure is

             P = F / A

in this case the force is equal to the weight of the body

              P = 14700 / 0.25

              P = 58800 Pa

onyebuchinnaji

(1) The area of the base of the block is 0.25 m².

(2) The volume of the block is 0.5 m³.

(3) The mass of the block is 1500 kg.

(4) The weight of the block is 14,700 N.

(5) The pressure exerted by the block is 58,800 N/m²

The given parameters;

  • density of the block, ρ = 3000 kg/m³
  • height of the block, h = 2 m
  • side of the base, b = 0.5 m

The area of the base of the block is calculated as follows;

[tex]A = 0.5 \ m \ \times 0.5 \ m\\\\A = 0.25 \ m^2[/tex]

The volume of the block is calculated as follows;

[tex]V = Ah\\\\V = 0.25 \times 2\\\\V = 0.5 \ m^3[/tex]

The mass of the block is calculated as follows;

[tex]\rho = \frac{m}{V} \\\\m = \rho V\\\\m = 3000 \times 0.5\\\\m = 1500 \ kg[/tex]

The weight of the block is calculated as follows;

[tex]W = mg\\\\W = 1500 \times 9.8\\\\W = 14,700 \ N[/tex]

The pressure exerted by the block is calculated as follows;

[tex]P = \frac{F}{A} \\\\P = \frac{14,700}{0.25} \\\\P = 58,800 \ N/m^2[/tex]

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