1. Air pollution is a problem<span> because it can cause damage to animals, trees, plants, crops and water sources in the environment. </span>Pollution<span> in the </span>air<span> causes </span>problems<span> for aviation because it reduces visibility, while also being responsible for damaging buildings and other structures.
2. </span><span>The air we breathe has a very exact chemical composition; 99 percent of it is made up of nitrogen, oxygen, water vapor and inert gases. Air </span>pollution occurs<span> when things that aren't normally there are added to the air. A common type of air </span>pollution<span>happens when people release particles into the air from burning fuels.
3. </span>Pollution prevention (P2) is any practice that reduces, eliminates, or prevents pollution at its source. ... Reducing the amount of pollution produced means less waste to control, treat, or dispose of. Less pollution means less hazards posed to public health and the environment.
4. Why is it so important to have clean air?
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The steel would expand by 4. 8 * 10^-3 cm
<h3>How to determine the linear expansion</h3>
The change in length ΔL is proportional to length L. It is dependent on the temperature, substance, and length.
Using the formula:
ΔL= α LΔT
where ΔL is the change in length L = 10cm
ΔT is the change in temperature = 60° - 20° = 40° C
α is the coefficient of linear expansion = 1.2 x 10^-5 °C
Substitute into the formula
ΔL = 
ΔL =
cm
Therefore, the steel would expand by 4. 8 * 10^-3 cm
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Given data:
* The extension of the steel wire is 0.3 mm.
* The length of the wire is 4 m.
* The area of cross section of wire is,

* The young modulus of the steel is,

Solution:
The young modulus of the steel in terms of the force and extension is,

where F is the force acting on the steel wire,, l is the original length of the wire, dl is the extension of the wire, and A is the area,
Substituting the known values,

Thus, the force which produce the extension of 0.3 mm of the steel wire is 31.5 N.
Answer:
Explanation:
velocity of first projectile after 3 s
v = u - gt
v = 49.4 - 9.8 x 3
= 20 m /s
Velocity of second projectile after 3 s after being dropped from rest
v = u + gt
= 0 + 9.8 x 3
= 29.4 m /s
They will be moving in opposite direction at the time of meeting , so their relative velocity
= 20 + 29.4 = 49.4 m /s
From the frame of reference of the first projectile, the velocity of the second projectile will be 49.4 m /s .