The speed of roller coaster car is about 9.4 m/s
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Further explanation
Centripetal Acceleration can be formulated as follows:
[tex]\large {\boxed {a = \frac{ v^2 } { R } }[/tex]
a = Centripetal Acceleration ( m/s² )
v = Tangential Speed of Particle ( m/s )
R = Radius of Circular Motion ( m )
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Centripetal Force can be formulated as follows:
[tex]\large {\boxed {F = m \frac{ v^2 } { R } }[/tex]
F = Centripetal Force ( m/s² )
m = mass of Particle ( kg )
v = Tangential Speed of Particle ( m/s )
R = Radius of Circular Motion ( m )
Let us now tackle the problem !
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Complete Question:
A roller coaster car is going over the top of a 18-m-radius circular rise. at the top of the hill, the passengers "feel light," with an apparent weight only 50% of their true weight. How fast is the coaster moving?
Given:
radius of circular motion = R = 18 m
weight of passengers = w
apparent weight of passengers = N = 50%w
Asked:
speed of roller coaster car = v = ?
Solution:
[tex]\Sigma F = m \frac{v^2}{R}[/tex]
[tex]w - N = m \frac{v^2}{R}[/tex]
[tex]w - 50\% w = m \frac{v^2}{R}[/tex]
[tex]50\% w = m \frac{v^2}{R}[/tex]
[tex]50\% mg = m \frac{v^2}{R}[/tex]
[tex]\frac{1}{2} g = \frac{v^2}{R}[/tex]
[tex]v^2 = \frac{1}{2} gR[/tex]
[tex]v = \sqrt { \frac{1}{2} g R }[/tex]
[tex]v = \sqrt { \frac{1}{2} \times 9.8 \times 18}[/tex]
[tex]v = \frac{21}{5} \sqrt{5} \texttt{ m/s}[/tex]
[tex]\boxed{v \approx 9.4 \texttt{ m/s}}[/tex]
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Learn more
- Impacts of Gravity : https://brainly.com/question/5330244
- Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454
- The Acceleration Due To Gravity : https://brainly.com/question/4189441
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Answer details
Grade: High School
Subject: Physics
Chapter: Circular Motion
