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(50) Soluzioni simulazione prova d'esame n. 3 • PHQWUHO¶XOWLPRYHUWLFHFKHLQGLFKLDPRFRQ/VLWURYLVXOODIDFFLD9$% 6H LQGLFKLDPR FRQ [ O¶DOWH]]D GL WDOH SDUDOOHOHSLSHGR H VH]LRQLDPR OD SLUDPLGH FRQ XQ SLDQR SHUSHQGLFRODUHDOODEDVHHSDVVDQWHSHU9HSHULOSXQWRPHGLR+GHOO¶LSRWHQXVD$%WDOHSLDQR SDVVHUjSHUODVLPPHWULDGHOODILJXUDDQFKHSHULOYHUWLFH/GHOSDUDOOHOHSLSHGR 3HUWDQWRGDOODVLPLOLWXGLQHGHLWULDQJROLUHWWDQJROLFKHVLYHQJRQRDIRUPDUHSRVVLDPRVFULYHUH. = CH : CV VF ⋅ CH 2 (h − x ) = 'DFXLVLULFDYDFKH FL = 2h CV. ODSURSRU]LRQH FL : VF. 0DVHRVVHUYLDPRFKH)/qODGLDJRQDOHGHOODEDVHGHOSDUDOOHOHSLSHGRLOODWR l GHOTXDGUDWRGL EDVHGHOSDUDOOHOHSLSHGRVLRWWHUUjGLYLGHQGRODGLDJRQDOHSHU. 2 l =. FL 2. =. h−x 2h. 2. 3HUWDQWRLOYROXPHGHOSDUDOOHOHSLSHGRVDUjGDWRGD V. §h−x· = x ⋅¨ ¸ © 2h ¹. 3HUWURYDUHSHUTXDOHYDORUHGL[LOYROXPHqPDVVLPRGHULYLDPR9RWWHQHQGR. V '=. h 2 − 4hx + 3 x 2 4h 2. 3RQHQGR 9¶! H ULVROYHQGR VL WURYD WHQHQGR FRQWR GHOOH FRQGL]LRQL JHRPHWULFKH FKH ULFKLHGRQR 0 <. x < h

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