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lawyer [7]
3 years ago
7

Two passengers are on a moving sidewalk at the airport. Passenger A is

Physics
1 answer:
Oksanka [162]3 years ago
8 0

Answer:

V = - 2[m/s]

Explanation:

This is a classic problem of relative velocities, in order to solve it we must understand each of the velocities, given in the initial data of the problem.

VA = velocity of passenger A, = 1.5[km/h], it is the same velocity of the sidewalk

VB = 2 + 1.5 = 3.5 [km/h], that is the velocity observed by a person outside from the sidewalk.

And from the perspective of passenger B the speed of passanger A is:

VA-VB = - 2 [m/s]

It means that the passenger B is seen how passenger A is getting close to him and then passed.

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Answer:

a) 633.39 J

b) 0.28

Explanation:

a)The kinetic energy of the player = \frac{1}{2} mv^{2}

Work done by friction = energy change of the player

                                    = \frac{1}{2} 67(4.35)^{2} = 633.9 J

b) Assuming the frictional force stays constant,

Work done by friction = Frictional force×distance

Frictional force = kinetic friction(μ)×normal reaction

Normal reaction = weight = mass×gravitational acceleration ( g=10m/s2 )

Combining these equations

633.9  =  F×3.4 ⇒ F = 186.44 N

F = μmg ⇒ μ = F/mg

                     = 186.44/670

                      = 0.28

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