Physics, 18.11.2020 16:50 neidaq12345

A volume of air increases 0.227 m^3 at a net pressure of 2.07 x 10^7 Pa. How much work is done on the air?

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Answer from: ivixlkoko

The work done on the air is 4.699 x 10⁶ Joules

Explanation:

Given;

increase in air volume, ΔV = 0.227 m³

net pressure of the air, P = 2.07 x 10⁷ Pa

The work done on the air is given by;

W = PΔV

Where;

W is the work done on the air

P is the net pressure

ΔV is the increase in air volume

Substitute the given values and solve for work done;

W = (2.07 x 10⁷ Pa) (0.227 m³)

W = 4.699 x 10⁶ Joules

Therefore, the work done on the air is 4.699 x 10⁶ Joules

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Answer from: Quest

electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. shielding is typically applied to enclosures to isolate electrical devices from their surroundings, and to cables to isolate wires from the environment through which the cable runs. electromagnetic shielding that blocks radio frequency electromagnetic radiation is also known as rf shielding.

the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia

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A volume of air increases 0.227 m^3 at a net pressure of 2.07 x 10^7 Pa. How much work is done on th...

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