please explain this!!
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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
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