The volume of a warmed part of the air is reduced and its density increases.
Explanation:
In a convective form of heat transfer, the volume of a warmed part of air is not reduced and its density does not increase.
During convection, heat causes the warm part of the air to expand and its volume increases. When volume increases, density is reduced.
- Convection is a form of heat transfer that involves the actual movement of particles of the medium.
- It usually occurs in fluids i.e gases and liquids.
- In convection, the cold part exerts a buoyant force on the warmer air below and causes it to rise.
- As the warmer part is rising the cooler part replaces it and a convective cell is formed in the process.
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The difference in the pressure between the inside and outside will be 369.36 N/m²
<h3>What is pressure?</h3>
The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.
It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.
The given data in the problem is;
dP is the change in the presure=?
Using Bernoulli's Theorem;

Hence, the difference in the pressure between the inside and outside will be 369.36 N/m²
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Using the formula: E = kQ / d² where E is the electric field, Q is the test charge in coulomb, and d is the distance.
E = kQ / d²
k = 9 x 10^9 N-m²/C²
Q = 6.4 x 10^-5 C
d = 2.5 x 10^-2 m
Substituting the given values to the equation, we have:
E = (9 x 10^9)(6.4 x 10^-5) / (2.5 x 10^-2) ²
Electric field at the test charge is 921600000 N/C
Answer:
b) false
Explanation:
Since in the given situation it is mentioned that the mass conversation would be simple and same applied to the fluids with care also the conservation of momentum would be applied to any type of closed system like collisions
But as we know that
In the inelastic collisions, the total kinetic energy would not be remain conserved
So the given statement is false