Question

a cubical block of weight 80N is kept on the levelled horizontal surface. If the side of the block is 4cm then pressure exerted by the block is​

Answer 1

Given:

A cubical block of weight 80N is kept on the levelled horizontal surface. The side of the block is 4cm .

To find:

Pressure exerted by object ?

Calculation:

• Area of one surface = side² = 4² = 16 cm².

• Now, 16 cm² = 16 × 10^(-4) m².

So, pressure will be :

$P = \dfrac{force}{area}$

$\implies \: P = \dfrac{80}{16 \times {10}^{ - 4} }$

$\implies \: P = 5 \times {10}^{4} \: Pascal$

So, pressure on surface is 5 × 10⁴ Pascal.

Answer 2

Answer:

Given:

A cubical block of weight 80N is kept on the levelled horizontal surface. The side of the block is 4cm .

To find:

Pressure exerted by object ?

Calculation:

Area of one surface = side² = 4² = 16 cm².

Now, 16 cm² = 16 × 10^(-4) m².

So, pressure will be :

P =

$\frac{force}{area}$

⟹P=

$\frac{80}{16 \times {10}^{ - 4} } </p><p>$

⟹P=5×10⁴ Pascal

So, pressure on surface is 5 × 10⁴ Pascal.