<u>Answer</u>:
The coefficient of static friction between the tires and the road is 1.987
<u>Explanation</u>:
<u>Given</u>:
Radius of the track, r = 516 m
Tangential Acceleration
= 3.89 m/s^2
Speed,v = 32.8 m/s
<u>To Find:</u>
The coefficient of static friction between the tires and the road = ?
<u>Solution</u>:
The radial Acceleration is given by,




Now the total acceleration is
=>
=>
=>
=>
The frictional force on the car will be f = ma------------(1)
And the force due to gravity is W = mg--------------------(2)
Now the coefficient of static friction is

From (1) and (2)


Substituting the values, we get


The density increases.
When gases are compressed, their volume decreases, and the resulting pressure increases. The temperature will change if either P or V are held constant. Since the volume decreases, then density, or m/V, increases.
P×V ~ T
Answer:
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