FTCose1. For each of the four trials, determine the airplane's mass fromEqn. (1)m =g.2. For each of the four trials,(a) Find the period of the airplane's motion, which we will call T.(Frsine) T(b) Find the airplane's mass fromEqn. (2)4nrIn Eqn. (2) r is the radius of the circular path3. The masses of the airplane (as measured on a scale) are 0.2935 kg for trial 13B0.2147kg for trial 2 ; 0.3935 kg for trial 3 ; 0.2939 kg for trial 4.4. (a) Find the percent difference between each of masses given in step 3 above and thecorresponding mass calculated using Eqn. (1).(b) Find the percent difference between each of masses given in step 3 above and thecorresponding mass calculated using Eqn. (2). TrialDiameter ofTension inAngle ofStringTime for TwoAirplane's PathStringRevolutions190.5 cm4.5 N522.5000 sec104.5 cm7.5741.7333 sec87.4 cm6.5522.3917 sec483.5 cm5.2502.3250 secDiameter for trial 1 is: 75.5 sin (52) = (45.31) (2) = 90.5 cmDiameter for trial 2 is: 54.5 sin (74) = (52.39) (2) = 104.5 cmDiameter for trial 3 is: 55.5 sin (52) = (43.73) (2) = 87.4 cmDiameter for trial 4 is: 54.5 sin (50) = (41.75) (2) = 83.5 cm

Since you have posted multiple questions with multiple sub parts, we will provide the solution only…

Trial 1 2 3 4 Diameter of Airplane's Path 90.5 cm 104.5 cm 87.4 cm 83.5 cm Tension in String 4.5 N 7.5 6.5 5.2 Angle of String Diameter for trial 1 is: 75.5 sin (52) = (45.31) (2) = 90.5 cm Diameter for trial 2 is: 54.5 sin (74) = (52.39) (2) = 104.5 cm Diameter for trial 3 is: 55.5 sin (52) = (43.73) (2) = 87.4 cm Diameter for trial 4 is: 54.5 sin (50) = (41.75) (2) = 83.5 cm 52 74 52 50 Time for Two Revolutions 2.5000 sec 1.7333 sec 2.3917 sec 2.3250 sec

Trial 1 2 3 4 Diameter of Airplane's Path 90.5 cm 104.5 cm 87.4 cm 83.5 cm Tension in String 4.5 N 7.5 6.5 5.2 Angle of String Diameter for trial 1 is: 75.5 sin (52) = (45.31) (2) = 90.5 cm Diameter for trial 2 is: 54.5 sin (74) = (52.39) (2) = 104.5 cm Diameter for trial 3 is: 55.5 sin (52) = (43.73) (2) = 87.4 cm Diameter for trial 4 is: 54.5 sin (50) = (41.75) (2) = 83.5 cm 52 74 52 50 Time for Two Revolutions 2.5000 sec 1.7333 sec 2.3917 sec 2.3250 sec

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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