Physics, asked by sonali8987, 9 months ago

A 5 kg wooden block is kept on a table top. If its
dimensions are 40 cm x 30 cm x 20 cm, find the
pressure exerted by the wooden block on the table
top, if it is made to lie on the table top with its
sides of dimensions
(i) 40 cm x 20 cm
(ii) 40 cm x 30 cm​

Answers

Answered by banerjeeatanu692
1

Answer:

Answer

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sharinkhan

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mass of the wooden block = m= 5 kgdimensions = 40 cm × 20 cm × 10 cm

the weight of the wooden block applies a thrust on the table top.That is,Thrust = F = mg= 5 kg × 9.8 m/s²= 49 N

Area of a side = length × breadth

= 20 cm × 10 cm

= 200 cm²

= 0.02 m²

Pressure = thrust/ area

Pressure = 49 /0.02= 2450 N/m²

The same thrust in implied when it lies on the given dimension.

Area= length × breadth= 40 × 20= 800= 0.08 m²

Pressure = 49/0.08= 612.5

Answered by TejeswiniSure
1

Explanation:

1. We know, Pressure = Force/ area

So, Pressure = 50000/40*20

= 50000/800

= 62.5 Pa

2. We know, Pressure = Force/ area

So, Pressure = 50000/40*30

= 50000/1200

= 41.67 Pa

I hope the answer is right.

Hope it helps u :)

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