find the ratio of the pressure exerted by a block of 400N when placed on a table top along its two different sides with dimensions 20 cm *10cm and 10 cm * 15 cm.
Answers
Given: Pressure is exerted by a block of 400N when placed on a table top along its two different sides with dimensions 20 cm *10cm and 10 cm * 15 cm.
To find: Ratio of pressure exerted
Solution: Pressure exerted by a body is equal to the the force applied to a body perpendicular to its area.
Here, Force = 400 N
Let the first surface have dimensions 20 cm and 10 cm.
Area of this surface
= length × breadth
= 20 × 10
= 200 cm^2
= 0.02 m^2 ( 1 m^2 = 10^-4 cm^2)
Pressure on this surface
= Force/Area
= 400 / 0.02
= 20000 Pa
Let the second surface have dimensions 10 cm and 15 cm.
Area of this surface
= length × breadth
= 10 × 15
= 150 cm^2
= 0.015 m^2 ( 1 m^2 = 10^-4 cm^2)
Pressure on this surface
= Force/Area
= 400 / 0.015 Pa
Ratio of pressure
= Pressure on first surface/ Pressure on second surface
= 20000/(400/0.015)
= 20000× 0.015/400
= 50 × 0.015
= 0.75
= 75/100
= 3/4
Ratio of pressure= 3:4
Therefore, the ratio of pressure exerted is 3:4.
Ratio of the pressure exerted is 3 : 4 by a block of 400N when placed on a table top along its two different sides with dimensions 20 cm *10cm and 10 cm * 15 cm.
Given:
- Block of 400 N
- Placed on table in different configuration
- 20 cm * 10 cm
- 10 cm * 15 cm
To Find:
- Ratio of Pressure exerted by Block in two cases
Solution:
Pressure = Force /Area
As Force is constant here hence Pressure is inversely proportional to Area
Step 1:
Calculate area in first case
A₁ = 20 * 10 = 200 cm²
Step 2:
Calculate area in second case
A₂ = 10 * 15 = 150 cm²
Step 3:
As Pressure is inversely proportional to Area
Hence Ratio will be
A₂ : A₁
150 : 200
15 : 20
3 : 4
Ratio of the pressure exerted is 3 : 4
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