Question

please explain this!!​

Answer 1

Answer:

Given:

Mass of the wooden block= 5 kg

Dimensions = 40 cm x 25 cm x 10 cm

The weight of the wooden block applies a thrust on the top of the table.

Thrust = F = m x g

                = 5 kg x 9.8 m s-2

                = 49 N

Considering the sides of dimensions, 

(a) 25 cm x 10 cm

(b) 40 cm x 25 cm

 

(a) When the block lies on its sides of dimensions 25 cm x 10 cm

 

Area of a side = length x breadth

                          = 25 cm x 10 cm = 250 cm2 = 0.025 m2

Therefore,

 

The pressure exerted by 25 cm x 10 cm is 1960 N m-2.

 

(b) When the block lies on its sides of dimensions 40 cm x 25 cm

 

Area of a side = length x breadth

                       = 40 cm x 25 cm = 1000 cm2 =  0.1m2

Therefore, pressure Exerted by 40 cm × 25 cm is

490N m-2.

HAPPY TO HELP