<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
=  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


 
        
             
        
        
        
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Thanks for sharing that information.  After extensive calculation, 
we can say with assurance that after some number of seconds, 
a loud "crunch" is perceived by the souls aboard the ill-fated vessel.
        
             
        
        
        
Anything that is made of atoms I believe. Matter is basically everything concrete that is not energy
        
             
        
        
        
Answer:
31.831 Hz.
Explanation:
<u>Given:</u>
The vertical displacement of a wave is given in generalized form as 

<em>where</em>,
- A = amplitude of the displacement of the wave.
- k = wave number of the wave =  
 = wavelength of the wave. = wavelength of the wave.
- x = horizontal displacement of the wave.
 = angular frequency of the wave = = angular frequency of the wave = . .
- f = frequency of the wave.
- t = time at which the displacement is calculated.
On comparing the generalized equation with the given equation of the displacement of the wave, we get,

therefore, 

It is the required frequency of the wave.