Hi friends!!!. If you know the answer plz reply. Describe any five everyday applications where the area over which a force applied is reduced in order to increase the pressure produced. If you know the answer to this question ☝️☝️then plz give me the answer
Answers
Answer:
tyres of vehicles
Explanation:
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Force, Pressure and Its Applications
Pressure is the force per unit area. The SI Unit of Pressure is the pascal (Pa). A pascal can be defined as a force of one newton applied over a surface area of one meter square. In other forms,
Pressure = Force / Area
The pressure is dependent on the area over which the force is acting, without any change in the force, the pressure can be increased and decreased. The Force applied to be constant if the surface becomes smaller the pressure increases and vice versa.
Some Practical Examples
Let us take an example of pressure: take into consideration a sharp needle, it has a small surface area. However, consider a pencil which is very blunt in the back. It has more surface area than the needle. If we poke the needle in our palm, it will definitely hurt us as the needle gets pierced inside our skin whereas if we poke the blunt side of the pencil into our hand it wouldn’t pain at all.
It is due to the fact that the area of contact between the palm and the needle is very small, further the pressure is in a way large. However, the area of contact between the pencil and the palm is more, therefore the pressure is less.
Another example we can take as – A brick sitting on a surface exerts a force equal to its weight on the object it is resting on. Now we know that a rectangular brick has a wide surface and thin surfaces on the sides. By changing the orientation of the brick resting on a surface, we are effectively changing the pressure acting on the surface by the same brick. You can refer to the image below for other information –
Pressure and its Application
By this we can conclude that the two factors that determine the magnitude of the pressure are:
Force: The force which states that the greater the force, the greater is the pressure and vice-versa.
Coverage Area: which states that the greater the area less is the pressure and vice-versa.
Browse more Topics under Mechanical Properties Of Fluids
Streamline Flow
Surface Tension
Viscosity
Bernoulli’s Principle and Equation
What is Bernoulli’s Principle and Equation?
Daily Life Pressure and Its Application
The application of atmospheric pressure, or in simple words, air pressure is referred to as the numerous activities that we do or observe every day without a realization of any kind of push or pressure. We are able to survive this high-pressure condition as we have evolved and somewhere managed our body functions withstand this amount of high pressure.
Solved Example for You on Pressure and Its Application
Question. What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20 °C) is 2.50 × 10–2 N m–1? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01 × 105 Pa).
Solution: Excess pressure inside the soap bubble is 20 Pa;
Soap bubble is of radius, r = 5.00 mm = 5 × 10–3 m
The surface tension of the soap solution, S = 2.50 × 10–2 Nm–1
A relative density of the soap solution = 1.20
Density of the soap solution, ρ = 1.2 × 103 kg/m3
Air bubble formed at a depth, h = 40 cm = 0.4 m
Radius of the air bubble, r = 5 mm = 5 × 10–3 m
1 atmospheric pressure = 1.01 × 105 Pa
Acceleration due to gravity, g = 9.8 m/s2
Hence, the excess pressure inside the soap bubble is:
P=4S/r
=(4×2.5×10-2)/5×10-3
=20Pa
Therefore, the excess pressure inside the soap bubble is 20 Pa. The excess pressure inside the air bubble is :
P’=2S/r
=(2×2.5×10-2)/5×10-3
=10Pa
Therefore, the excess pressure inside the air bubble is 10 Pa. At a depth of 0.4 m, the total pressure inside the air bubble,
Atmospheric pressure + hρg + P’
=1.01×105+0.4×1.2×103x9.8+10
=1.057×105Pa
=1.06×105Pa
Therefore, the pressure inside the air bubble is 1.06×105 Pa.