<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
It’s D. An enlargement (hope this helps!)
Answer:
U = (ε0AV^2) / 2d
Explanation:
Where C= capacitance of the capacitor
ε0= permittivity of free space
A= cross sectional area of plates
d= distance between the plates
V= potential difference
First, the capacitance of a capacitor is obtained by:
C = ε0A/d.
Starting at the formula , U= (CV^2)/2. Formula for energy stored in a capacitor
Substitute in for C:
U = (ε0A/d) * V^2 / 2
Hence:
U = (ε0AV^2) / 2d
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