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Pie
3 years ago
13

You're driving your new sports car at 80 mph over the top of a hill that has a radius of curvature of 540 m. What fraction of yo

ur normal weight is your apparent weight as you crest the hill?
Physics
1 answer:
luda_lava [24]3 years ago
4 0

Answer:

75.84%

Explanation:

We were given Speed of the sports car, v as 80 mph , we can convert to m/s for unit consistency.

v=80mph= 35.76 m/s

The radius of curvature is given as , r = 540 m

✓ the normal weight can be denoted as Wn

✓ the apparent weight of the person can be denoted as Wa

Wn= normal weight= mg

Wa=apparent weight = (mg - mv^2/r)

g= acceleration due to gravity= 9.8m/s^2

The apparent weightand normal weight has a ratio of

Mn/Ma= [mg - mv^2/r]/mg ........eqn(1)

If we simplify eqn(1) we have

Mn/Ma=[g - vr^2/g].............eqn(2)

Then substitute the given values

Mn/Ma=9.8 - [(35.76^2)/540]/ 9.8

=0.758×100%

Mn/Ma=75.84%

Hence, the required fraction is 75.84%

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