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Fantom [35]
3 years ago
13

If the maximum acceleration that is tolerable for passengers in a subway train is 1.21 m/s2 and subway stations are located 810

m apart, what is the maximum speed a subway train can attain between stations
Physics
1 answer:
Rasek [7]3 years ago
7 0

Answer:

v = 44.27 m / s

Explanation:

This is an exercise in kinematics since they ask the maximum speed of the train

           v² = v₀² + 2 a x

a train departs with zero initial velocity

           v² = 2 a x

           v = √ 2ax

calculate

           v = √ (2 1.21 810)

            v = 44.27 m / s

in this solution we have assumed that the train uses the entire distance with acceleration

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As a box slides down a ramp, friction does 23.0 joules of work. At the bottom of the ramp, the box has 3.8 joules of kinetic ene
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The high of the ramp is 2.81[m]

Explanation:

This is a problem where it applies energy conservation, that is part of the potential energy as it descends the block is transformed into kinetic energy.

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E_{k}=0.5*m*v^{2}\\\\where:\\E_{k}=3.8[J]\\v = 2.8[m/s]\\m=\frac{E_{k}}{0.5*v^{2} } \\m=\frac{3.8}{0.5*2.8^{2} } \\m=0.969[kg]

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m*g*h + W_{f}=3.8\\ 0.969*9.81*h - 23= 3.8\\h = \frac{23+3.8}{0.969*9.81}\\ h = 2.81[m]

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