Question

A force of (2500+5)N is applied over an area of (0.32┬▒0.02) m^2. Calculate the pressure enerted

Answer 1

your question is -> A force of (2500 ± 5)N is applied over an area of (0.32 ± 0.02) m^2. calculate the pressure exerted.

we know, pressure is the force applied per unit area in a surface.

i.e., pressure = force applied/area

P = F/A

here, F = 2500N , A = 0.32m²

so, P = 2500/0.32 = 7,812.5N/m²

now using formula ∆P/P = ∆F/F + ∆A/A , to find error of pressure .

here, P = 7812.5 , ∆F = 5, F = 2500, ∆A = 0.02m² and A = 0.32

so, ∆P/7812.5 = 5/2500 + 0.02/0.32

∆P = 7812.5(1/500 + 1/16) = 503.90625 ≈ 504

hence, appropriate value of pressure = (7812.5 ± 504) N/m²

Answer 2

Answer :-

your question is -> A force of (2500 ± 5)N is applied over an area of (0.32 ± 0.02) m^2. calculate the pressure exerted.

we know, pressure is the force applied per unit area in a surface.

i.e., pressure = force applied/area

P = F/A

here, F = 2500N , A = 0.32m²

so, P = 2500/0.32 = 7,812.5N/m²

now using formula ∆P/P = ∆F/F + ∆A/A , to find error of pressure .

here, P = 7812.5 , ∆F = 5, F = 2500, ∆A = 0.02m² and A = 0.32

so, ∆P/7812.5 = 5/2500 + 0.02/0.32

∆P = 7812.5(1/500 + 1/16) = 503.90625 ≈ 504

hence, appropriate value of pressure = (7812.5 ± 504) N/m²