Answer:
(a) 
(b) 
Explanation:
<u>Given:</u>
= The first temperature of air inside the tire = 
= The second temperature of air inside the tire = 
= The third temperature of air inside the tire = 
= The first volume of air inside the tire
= The second volume of air inside the tire = 
= The third volume of air inside the tire = 
= The first pressure of air inside the tire = 
<u>Assume:</u>
= The second pressure of air inside the tire
= The third pressure of air inside the tire- n = number of moles of air
Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.
Using ideal gas equation, we have

Part (a):
Using the above equation for this part of compression in the air, we have

Hence, the pressure in the tire after the compression is
.
Part (b):
Again using the equation for this part for the air, we have

Hence, the pressure in the tire after the car i driven at high speed is
.
.The path of a celestial body or an artificial satellite as it revolves around another body due to their mutual gravitational <span>attraction.</span>
Answer:
20.25 m
Explanation:
- <u>Centripetal acceleration </u>is given by; the square of the velocity, divided by the radius of the circular path.
That is;
<em><u>ac = v²/r</u></em>
<em> </em><em><u> Where; ac = acceleration, centripetal, m/s², v is the velocity, m/s and r is the radius, m</u></em>
Therefore;
r = v²/ac
= 27²/36
= 20.25 m
Hence the radius is 20.25 meters
2a for example the first one,2sec. You know that every second it moves 3metres further. So 2x3=6 but you start at 0.50m so 6+0.50=6.5