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(answered) – 1″ steel on 1″ steel (66.68 g each) Scale 2.38 : 1, 0.00287

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(answered) – 1″ steel on 1″ steel (66.68 g each) Scale 2.38 : 1, 0.00287DescriptionSolution downloadThe QuestionPhysics collision theory. Please take a look at the attachments. It shows 5 balls. We need to get the real distances by a scale factor of 2.53 cm/in. Each ball represents 10 frames. We need to get both distance and time which is velocity.m1v1x(initial)= m1v1xfinal + m2v2xfinal = m1v1finalcostheta1+ m2v2finalcostheta2m1v1yinitial=m1v1yfinal + m2v2finaly =m1v1final sintheta1 + m2v2final sin theta2Initial projectile angle = 0 degreesIf each projectile rolled down a ramp from a ht of 34 cm, what would be the predicted KE (rolling, not sliding without friction)What could account for any discrepancy?KE for a rolling sphere is 0.7 mv2(For all 5) The lines are on the picture, but we still have to examine their properties to find their lengths which will have to be converted to their real values and angles.ERROR ANALYSIS!Calculate percentage error and can the error come from measurement errors or something else happening…1? steel on 1? steel (66.68 g each)Scale 2.38 : 1, 0.00287 sec/frameRed line ? scales to 14.7 cm, 30 frames 1.7 m/s -1 0Green line ? scales to 17.3 cm, 70 frames 0.86 m/s +44 +45Blue line ? scales to 11.0 cm, 30 frames 1.28 m/s-34 -331? Steel (66.68 g) 1? Poly ( 7.39g)Scale 2.43:1, 0.00291 sec/frameRed Line ? scales to 13.9 cm, 30 frames 1.59 m/s -2 0Green Line ? scales to 20.7 cm, 40 frames 1.78 m/s 28 30Blue Line ? scales to 17.8 cm, 40 frames 1.53 m/s -3 -11? Steel (66.68 g) 1? Wood (5.06 g)Scale 2.45:1, 0.00287 sec/frameRed line ?Green line ?Blue line ?0.75? steel (28.2 g) on 1? steel (66.68 g)Scale 2.49 : 1, 0.00287 sec/frameRed line ?Green line ?Blue line ?1? steel (66.68 g) on 1? PTFE (18.66 g)Scale :1, 0.00303 sec/frame

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