<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


you can subtract the atomic number from the mass number to find the number of neutrons.
Originally there must been
1,4775E6 + 2.25E4 = 147.75E4 + 2.25E4 = 150E4 present at start
% = 2.25 / 150 = 1.5 % of 235 U left
When drinking at a private event, you should assume that drinks will be STRONGER THAN NORMAL.
At private events, some hosts have the habit of mixing different drinks together in order to increase the intoxicating power of the drinks. This does not normally happen when one is buying from restaurants or other commercial places. Thus, to be on the safe side, one should always assume that drinks will be stronger when one is attending a private event, this will caution one to drink responsibly in order to avoid intoxication.
Answer:
5.33*10^-3 seconds
Explanation:
c = d/t
c = speed of light constant (3.0*10^5 km/s)
d = distance (1600 km)
t = ?
3.0*10^5 = 1600/t
t = 1600/3.0*10^5
t = 5.33*10^-3 seconds
I hope this helped! :)