Physics: A cyclist coasts up a 9.00° slope, traveling 12.0 m along the road to the top of the hill. If the cyclist’s initial speed is 9.00 m/s, what is the ﬁnal speed? Ignore friction and air resistance.

A cyclist coasts up a 9.00° slope, traveling 12.0 - id9001205824418, myahprovenzano

july00

22.01.2022

$v_{f} \approx 6.647\,\frac{m}{s}$

Explanation:

The final speed of the cyclist is determined by applying the Principle of Energy Conservation:

$\frac{1}{2}\cdot m\cdot v_{o}^{2} + m\cdot g\cdot h_{o} = \frac{1}{2}\cdot m\cdot v_{f}^{2} + m\cdot g\cdot h_{f}$

$\frac{1}{2} \cdot v_{o}^{2} + g\cdot (h_{o}-h_{f}) = \frac{1}{2}\cdot v_{f}^{2}$

$v_{f}^{2}=v_{o}^{2} + 2\cdot g \cdot (h_{o}-h_{f})$

$v_{f} = \sqrt{v_{o}^{2}+2\cdot g \cdot (h_{o}-h_{f})}$

$v_{f} = \sqrt{v_{o}^{2}-2\cdot g \cdot \Delta s \cdot \sin \theta}$

$v_{f} = \sqrt{(9\,\frac{m}{s} )^{2}-2\cdot (9.807\,\frac{m}{s} )\cdot (12\,m)\cdot \sin 9^{\textdegree}}$

$v_{f} \approx 6.647\,\frac{m}{s}$

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hahalol123goaway

28.09.2022

a) momentum of first car = 23159.72 kg.m/s

b) velocity of second car = 10.64 m/s

Explanation:

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