Please check if I solved this question correctly and if the answer is correct A car of mass 4540 kg is travelling over a hill with a radius of curvature of 13.9 m, ata speed of 8.5 m/s. What is the magnitude of the normal force acting on the carwhen the car is at the very top of the hill?IS Mass of the car(m)4540 kg- The radius of curvature of the hill (r) = 13.9.-Speed of thecar v = 8.5 m/sthe magnitude of the normal forceactingGiven dataFindingAt the top of theFnet-Nmg-N = my ²mgmg(mye is the centripetal force in circular motion, mg istheweightmy²ORthe hill, the netforce²² = NN =mg4542= 4540 kg x 9.8 m/s² =(9.8 m/s² -= 4540 kgmy2on the car,4540 kg (8.5m/s)?(8.5 m/s2)13.9m2- 2(9.8. m/s² - 5. 197 ms.= 4540 kg= 4540 kg x 4.603 m/s²=20897.62 N

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A car of mass 4540 kg is travelling over a hill with a radius of curvature of 13.9 m, at a speed of 8.5 m/s. What is the magnitude of the normal force acting on the car when the car is at the very top of the hill?

A car of mass 4540 kg is travelling over a hill with a radius of curvature of 13.9 m, at a speed of 8.5 m/s. What is the magnitude of the normal force acting on the car when the car is at the very top of the hill?

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter5: More Applications Of Newton’s Laws

Section: Chapter Questions

Problem 4OQ: An office door is given a sharp push and swings open against a pneumatic device that slows the door...

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Concept explainers

Centripetal Force

A particle, body, object or certain mass while travelling in a rotational path or semicircular path, experiences an inertia force. If that inertia force pulls the towards the inward of the imaginary circle in the rotation, it is termed as centripetal force.

Question

Please check if I solved this question correctly and if the answer is correct

A car of mass 4540 kg is travelling over a hill with a radius of curvature of 13.9 m, at
a speed of 8.5 m/s. What is the magnitude of the normal force acting on the car
when the car is at the very top of the hill?
IS

Mass of the car
(m)
4540 kg
- The radius of curvature of the hill (r) = 13.9.
-Speed of the
car v = 8.5 m/s
the magnitude of the normal force
acting
Given data
Finding
At the top of the
Fnet
-N
mg
-N = my ²
mg
mg
(mye is the centripetal force in circular motion, mg is
the
weight
my²
OR
the hill, the netforce
²² = N
N =
mg
454
2
= 4540 kg x 9.8 m/s² =
(9.8 m/s² -
= 4540 kg
my2
on the car,
4540 kg (8.5m/s)?
(8.5 m/s2)
13.9m
2
- 2
(9.8. m/s² - 5. 197 ms.
= 4540 kg
= 4540 kg x 4.603 m/s²
=20897.62 N

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