1. A customer sits in an amusement park ride in which the compartment is to be p

Question

1. A customer sits in an amusement park ride in which the compartment is to be pulled downward in the negative direction of a y axis with an acceleration magnitude of 1.24g, with g = 9.80m/s2. A 0.567 g rest on the customer's knee. Once the motion beginns and the unit-vector notation, what is the coin's acceleration relative to (a) the ground and (b) the customer? (c) How long does the coin take to reach the compartment ceiling, 2.20 m above the knee? In unit-vector notation, what are (d) the actual force on the coin and (e) the apparent force according to the customer's measure of the coin's acceleration?

Explanation / Answer

Given :     Acceleration of the compartment = -g jm/s2 a) Acceleration of the coin w.r.t ground = -gj m/s2 [Since it will freelyfall] = -9.8 j m/s2 b) Acceleration of the coin w.r.t customer = (1.24 - 1)gj m/s2 = 0.24g j m/s2 = 0.24g j m/s2 = 2.35 j m/s2 c) Time the coin takes to reach the ceiling of the compartmentat a dsistance s = 2.2m above the customer's knee, s = ut + 1/2at2 2.20m = 0 + (0.5)(2.35m/s2) t2 t = 1.37s d) mass = 0.567 x 10-3kg Actual force on the coin,F =mg F = -( 0.567x10-3kg)( 9.8m/s2)j   =-5.56x10-3jN   =-5.56x10-3jN    e) Apparent force according to customersmeasure of the acceleration is,     F =(0.567 x10-3N)(2.35) jN       =1.33x10-3jN      I hope it helps you      I hope it helps you

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