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Because the net radial force is acting as the centripetal force, we can equate them and solve for v: Fradial, net = Fc mv? T-mg cos(45 )= v = (T-mg cos(45')) -T-rg cos(45')

Because the net radial force is acting as the centripetal force, we can equate them and solve for v: Fradial, net = Fc mv? T-mg cos(45 )= v = (T-mg cos(45')) -T-rg cos(45')

College Physics
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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Why does the m change to an r in the rooted formulas. I want to understand why please.
回 X a762645ccccS:centripetal-acceleration-and-centrip... & * First attempt Taia, net Because the net radial force is acting as the centripetal force, we can equate them and solve for v: Fradial, net = F. mv² T-mg cos(45) v = (T-mg cos(45')) m -T - rg cos(45) NA 2.5 m V 3= 1.5 kg (32 N) - (2.5 m) (9.8 ) cos(45) m U 6.001 6.00 4/4 The speed of the mass is 6.00 5:47 PM 63°F E O 4) ENG 1/16/2022
Transcribed Image Text:回 X a762645ccccS:centripetal-acceleration-and-centrip... & * First attempt Taia, net Because the net radial force is acting as the centripetal force, we can equate them and solve for v: Fradial, net = F. mv² T-mg cos(45) v = (T-mg cos(45')) m -T - rg cos(45) NA 2.5 m V 3= 1.5 kg (32 N) - (2.5 m) (9.8 ) cos(45) m U 6.001 6.00 4/4 The speed of the mass is 6.00 5:47 PM 63°F E O 4) ENG 1/16/2022
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