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UBA - CienciaS Foro de Alumnos 2009-07-06T13:20:19-03:00 http://www.ubacs.com.ar/ubacs/feed.php?f=13&t=282 2009-07-06T13:20:19-03:00 2009-07-06T13:20:19-03:00 http://www.ubacs.com.ar/ubacs/viewtopic.php?t=282&p=7411#p7411 <![CDATA[Re: [PEDIDO] SEGUNDOS PARCIALES]]>
Por otra parte,si te interesan las "formulitas"...las coordenadas del CENTRO DE MASA DE UNA SUPERFICIE PLANA
se pueden calcular como..

con ,la funcion densidad de masa y D la region a integrar
y la coordenada y del centro de masa la obtenees cambiando x por y en el numerador del cociente anterior...

ademas,las coordenadas del CENTRO DE MASA DE UNA REGION TRIDIMENS. son:



con ,la funcion densidad de masa y D la region a integrar


a las coordenadas y ,z ,las obtenes analogamente,reemplazando esa x del numerador por y o por z segun la coordenada que desees hallar.

espero que les sirva de algo y tengan cuidado: fue escrito por mi :? ajja,en fin saludos



EDIT: ahora que me doy cuenta Joaquin habia puesto la def arriba de C.M ....discuplame Joaquin!!! no fue queriendo,es que ,como habian pregutado por "formulitas",supuse que estas no estaban y no me percate de mirar arriba :(,i.e:vi solo el fin d la pagina 1 :(

Estadísticas: Publicado por Maca.... — 06 Jul 2009, 13:20


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2009-06-29T14:15:07-03:00 2009-06-29T14:15:07-03:00 http://www.ubacs.com.ar/ubacs/viewtopic.php?t=282&p=7132#p7132 <![CDATA[Re: [PEDIDO] SEGUNDOS PARCIALES]]> . Veremos como será el sábado la cosa... Un slaudo

Estadísticas: Publicado por Erizo — 29 Jun 2009, 14:15


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2009-06-23T13:39:43-03:00 2009-06-23T13:39:43-03:00 http://www.ubacs.com.ar/ubacs/viewtopic.php?t=282&p=7084#p7084 <![CDATA[Re: [PEDIDO] SEGUNDOS PARCIALES]]>
era lo q necesitaba para abrir un poco mi mente

saludos!

Estadísticas: Publicado por inesi — 23 Jun 2009, 13:39


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2009-06-23T01:06:17-03:00 2009-06-23T01:06:17-03:00 http://www.ubacs.com.ar/ubacs/viewtopic.php?t=282&p=7080#p7080 <![CDATA[Re: [PEDIDO] SEGUNDOS PARCIALES]]> Estadísticas: Publicado por FJL — 23 Jun 2009, 01:06


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2009-06-22T16:59:13-03:00 2009-06-22T16:59:13-03:00 http://www.ubacs.com.ar/ubacs/viewtopic.php?t=282&p=7075#p7075 <![CDATA[Re: [PEDIDO] SEGUNDOS PARCIALES]]>




Estadísticas: Publicado por Joaquín — 22 Jun 2009, 16:59


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2009-06-22T14:41:01-03:00 2009-06-22T14:41:01-03:00 http://www.ubacs.com.ar/ubacs/viewtopic.php?t=282&p=7071#p7071 <![CDATA[Re: [PEDIDO] SEGUNDOS PARCIALES]]> O sea, sí, es el lugar donde se equilibra las masas supongo, pero la pregunta es si hay alguna formula o algo porque no se ni como plantear el ejercicio...
Por otro lado, alguien que me ayude a encaminar el 3?
Desde ya, gracias

Estadísticas: Publicado por Erizo — 22 Jun 2009, 14:41


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2009-06-19T19:47:11-03:00 2009-06-19T19:47:11-03:00 http://www.ubacs.com.ar/ubacs/viewtopic.php?t=282&p=7042#p7042 <![CDATA[Re: [PEDIDO] SEGUNDOS PARCIALES]]> Estadísticas: Publicado por Quimey — 19 Jun 2009, 19:47


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2009-06-19T10:45:06-03:00 2009-06-19T10:45:06-03:00 http://www.ubacs.com.ar/ubacs/viewtopic.php?t=282&p=7033#p7033 <![CDATA[Re: [PEDIDO] SEGUNDOS PARCIALES]]>
gracias!

Estadísticas: Publicado por inesi — 19 Jun 2009, 10:45


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2008-07-13T16:14:22-03:00 2008-07-13T16:14:22-03:00 http://www.ubacs.com.ar/ubacs/viewtopic.php?t=282&p=1374#p1374 <![CDATA[Re: [PEDIDO] SEGUNDOS PARCIALES]]> m2=80p2
mtotal=80p1+80p2
Imponiendo la condiciòn de que y del centro de masa fuera 20, saco que p/(p1+p2)=5/8
Luego las coordenadas del centro de masa son:(50/3, 20)
Siendo p la densidad de las dos placas juntas.

Es probable que no estè bien, aguardo comentarios

Estadísticas: Publicado por Mar — 13 Jul 2008, 16:14


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2008-07-13T00:38:16-03:00 2008-07-13T00:38:16-03:00 http://www.ubacs.com.ar/ubacs/viewtopic.php?t=282&p=1352#p1352 <![CDATA[Re: [PEDIDO] SEGUNDOS PARCIALES]]> Estadísticas: Publicado por Yossarian — 13 Jul 2008, 00:38


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