SHOW ALL WORK PLEASE (80%) Tue 4:32 PM Yara Fakhoury Q. Chrome File Edit View Hi
ID: 2120408 • Letter: S
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(80%) Tue 4:32 PM Yara Fakhoury Q. Chrome File Edit View History Bookmarks Window Help e O O LON CAPA p 07 C https:// s2 e msu.edu doboa2 64b4c8 59e Sc730fb4ff3019... a E oblems true&isbn; &search; A+mass of 4.95 kg+... Q Messages Courses Help Logout Since that will mean e 90 which is not possible since in that c Yara Nidal Fakhoury Student section: 731 PHY 183B Summer 2013 Main Menu Course Contents Course Contents» Correction 2 (07/17 We 11:59 PM Notes M Bookmark Evaluate Communicate Print A mass of 4.95 kg is suspended from a 1.55 m long string. It revolves in a horizontal circle as shown in the fiqure ing. It revolves in a horizontal c as shown in the ong S ween the string and the vertical n2 1.55 sin A T sin A where T ension in string; A angle the 1.55 sin A JT sin A 90.7 Tsin A2 A (1) resolving vertically, T cos A1 sin A2 A 1.06 cos A J sinA2 A 1.06 cos A 1 cos n2 A c If the string makes an angle of o 58.3 with the vertical, then what is the net force acting on the u should get the answer ass? Submit Answer Incorrect. Tries 2/12 Previous Tries What is the period of the pendulum e. the time it takes for the mass to complete one whole circle? Submit Answer Tries 0/12 ing. It revolves in a horizontal c as shown in the ong S Preferences on what is marked as Threaded View Chronological View Sorting/Filtering NEW een the string and the vertical. 4.31x101 deg options Export? Mark NEW posts no longer new NEW Anonymous 1 Reply (Sun Jul 14 12:31:21 am 2013 (EDT)) any suggestions? Show All media-92f-92171781... pngExplanation / Answer
Net force is in the horizontal direction providing centripetal acceleration as verical force is balanced..
Vertical direction
Tcos(58.3) = (4.95)(9.8)
T = 92.31N
Net force = Tsin(58.3) = 78.538N
78.538 = 4.95(w^2)*1.55sin(58.3)
w=3.46
T = 2pie/w = 1.81 secs
Hope i Could help
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