Web20 sep. 2024 · To find the torque by the force F= (2i+3j-5k) acting on r1= (2i+4j+7k) m about r2= (i+2j+3k) m We will find the torque at r1 and subtract it from the torque at r2 as individually they are torques about the origin. Using determinant method, τ1 =r1 × F = i (21+20) + j (-10 -14) + k (8-6) = (41i -24j + 2k) WebI_djupaste_h-herrar_Gothmogd3Q‘d3Q‘BOOKMOBI Õë ¨ ˜ Ú )¾ 2ª ;\ E N« X€ aâ k• uR ˆÜ ’q œ ¥¬"¯©$¹Y&Ân(Ëÿ*Ö6,à .ê 0ó 2üµ4 :6 ù8 è: #ä -{> 6á@ ?ÜB I¯D S7F ]NH f›J pZL z0N „ P †R — T âV ªÌX ´EZ ¾ \ Ç^ Ñ ` ×Âb Ý®d ä[f êôh ð¢j ö l û»n (p ;r at Jv ßx §z ¨ ”~ h€ ”‚ å(„ èä† øðˆ œŠ ŒŒ Ž ô !ä’ %Ô” )È ...
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Web1. If F=2i+3 +4k acts on a body and displaces it by $ = 3i+2j+5k , then the work done by the force is 1) 12 J 2) 20 J 3) 32 J 4) 64 J 2 Afon120Nationsten Solution Verified by Toppr Was this answer helpful? 0 0 Similar questions A weightlifter lifts a weight off the ground and … Web16 feb. 2024 · A force $(3i + 4j)$ Newton acts on a body and displaces it by $(3i + 4j)$ meters. The work done by this force is:A. 10 JB. 5 JC. 25 JD. 12 J. Ans: Hint: In the question, we need to determine the work done by the force to displace the body. For this, ... thentia toronto
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WebLawÅnforcementÎews (€0Ùork,Î.Y.)…à2 xol liöalu‚ 1‚ aæilepos=…y…ˆ336 ‚·‚·‚·‚·a„ /li€1‚â/…°„¸…ç…↹-list"èidden ... Web10 aug. 2024 · Force acting on a particle is 2i+3j work done by this force is zero when a particle is moved on a line 3y+kx=5 find value of k. force acting on a particle is 2i+3j work done by this force is zero when a particle is moved on a line 3y+kx=5 find value of k. 2 Answer(s) Answer Now. 1 Likes; 2 Comments; 0 Shares; Likes; http://ia-petabox.archive.org/download/manualofoccultis01seph/manualofoccultis01seph.mobi the ntintili factory