Respuesta :
Answer:
1) The pressure exerted by the metal box on the floor is 33.[tex]\overline 3[/tex] Pa
2) The pressure exerted at the base is 250 Pa
3) The pressure exerted at the point where the pen pushes against the paper is 24 MPa
Explanation:
1) Pressure is given by the following relation
[tex]Pressure, P = \dfrac{Force, F}{Area, A}[/tex]
[tex]\therefore P = \dfrac{F}{A}[/tex]
The force exerted by the weight of the metal box, F = 20 N
The dimensions of the side on which the metal box rests = 1 m by 0.6 m
∴ The area of the side on which the metal box rests, A = 1 m × 0.6 m = 0.6 m²
The pressure exerted by the metal box on the floor, 'P' is therefore given as follows;
[tex]\therefore P = \dfrac{F}{A} = \dfrac{20 \, N}{0.6 \, m^2} = \dfrac{100}{3} \ Pa = 33.\overline 3 \ Pa[/tex]
The pressure exerted by the metal box on the floor, P = 33.[tex]\overline 3[/tex] Pa
2) The dimensions of the rectangular container are;
Base dimension = 50 cm by 30 cm = 0.5 m by 0.3 m
The depth to which the water is filled in the container = 5 cm = 0.05 m
The density of water, [tex]\rho_w[/tex] = 1,000 kg/m³
The volume of water in the container, V = Base area × Height = 0.5 m × 0.3 m × 0.05 m = 0.0075 m³
∴ V = 0.0075 m³
The mass of the water, m = V × [tex]\rho_w[/tex] = 0.0075 m³ × 1,000 kg/m³ = 7.5 kg
m = 7.5 kg
The weight of the water, W = m × g = 7.5 kg × 10 m/s² = 75 N
The base area, A = 0.5 m × 0.6 m = 0.3 m²
The pressure exerted at the base, P = W/A = 75 N/(0.3 m²) = 250 Pa
The pressure exerted at the base, P = 250 Pa
3) The area of the point of the ball-pointed pen pushing against the paper, A = 1 mm² = 0.000001 m²
The force with which the point on the pen pushes against the paper = 24 N
The pressure exerted at the point where the pen pushes against the paper, P = W/A
∴ P = 24 N/(0.000001 m²) = 24,000,000 Pa
The pressure exerted at the point where the pen pushes against the paper, P = 24,000,000 Pa = 24 MPa.
