Question

Calculate the pressure exerted by brick which applies force of 2.5 N when a) it is placed upright on soil b) when it is placed on its widest base. The dimensions of the brick are 25 cm*10 cm* 5 cm​

Answer 1

Answer:

Answer of (a)

is 500 Pa and that of (b) is 100 Pa.

Explanation:

Thrust(Compressive force) = 2.5 N .

a).

Area of cross-section = {(5cm * 10 cm) / (100*100)} m².

= 1 / 200 m²

Therefore, Pressure =

$\frac{force \: exerted}{area \: of \: cross \: section}$

= 2.5N / (1/200) m².

= 500 Nm.

= 500 Pa.

b).

Area of cross-section = {(25cm * 10 cm) / (100*100)} m².

= 1 / 40 m²

Therefore, Pressure =

$\frac{force \: exerted}{area \: of \: cross \: section}$

= 2.5N / (1/40) m².

= 100 Nm.

= 100 Pa.

Answer 2

Explanation:

Answer:

Answer of (a)

is 500 Pa and that of (b) is 100 Pa.

Explanation:

Thrust(Compressive force) = 2.5 N .

a).

Area of cross-section = {(5cm * 10 cm) / (100*100)} m².

= 1 / 200 m²

Therefore, Pressure =

\frac{force \: exerted}{area \: of \: cross \: section}

areaofcrosssection

forceexerted

= 2.5N / (1/200) m².

= 500 Nm.

= 500 Pa.

b).

Area of cross-section = {(25cm * 10 cm) / (100*100)} m².

= 1 / 40 m²

Therefore, Pressure =

\frac{force \: exerted}{area \: of \: cross \: section}

areaofcrosssection

forceexerted

= 2.5N / (1/40) m².

= 100 Nm.

= 100 Pa.