Explanation:
It is given that,
Speed of the sports car, v = 85 mph = 37.99 m/s
The radius of curvature, r = 525 m
Let [tex]W_N[/tex] is the normal weight and [tex]W_A[/tex] is the apparent weight of the person. Its apparent weight is given by :
[tex]W_A=mg-\dfrac{mv^2}{r}[/tex]
So, [tex]\dfrac{W_A}{W_N}=\dfrac{mg-\dfrac{mv^2}{r}}{mg}[/tex]
[tex]\dfrac{W_A}{W_N}=\dfrac{g-\dfrac{v^2}{r}}{g}[/tex]
[tex]\dfrac{W_A}{W_N}=\dfrac{9.8-\dfrac{(37.99)^2}{525}}{9.8}[/tex]
[tex]\dfrac{W_A}{W_N}=0.719[/tex]
or
[tex]\dfrac{W_A}{W_N}=71.9\%[/tex]
Hence, this is the required solution.
The apparent weight is 71% of the normal weight while going through the crest of hill.
Given data:
The speed of car at top of hill is, v = 85 mph = 37.99 m/s.
The radius of curvature of path is, r = 525 m.
The expression for the apparent weight of car moving in a curved path is,
[tex]W_{a}=W_{n}-F_{c}[/tex]
Here, [tex]F_{c}[/tex] is the centripetal force and [tex]W_{n}[/tex] is the normal weight.
Solving as,
[tex]\dfrac{W_{a}}{W_{n}} =1-\dfrac{F_{c}}{W_{n}}\\\dfrac{W_{a}}{W_{n}} =1-\dfrac{\dfrac{mv^{2}}{r}}{mg}\\\dfrac{W_{a}}{W_{n}} =1-\dfrac{\dfrac{v^{2}}{r}}{g}\\\dfrac{W_{a}}{W_{n}} =1-\dfrac{\dfrac{37.99^{2}}{525}}{9.8}\\\dfrac{W_{a}}{W_{n}} =0.71\\[/tex]
Thus, we can conclude that the apparent weight is 71% of the normal weight while going through the crest of hill.
Learn more about the apparent weight here:
https://brainly.com/question/19410229?referrer=searchResults
