Asked by G. Jonathan

A uniform metre rule of Mass 120g is pivoted at 60cm Mark. At what point on metre rule should a mass of 50g be suspended for it to balance horizontally
Answered by R_scott
the 120 g mass acts at the center of the metre rule (50 cm mark)
... 10 cm from the pivot

the 50 g mass needs to be on the other side of the pivot (> 60 cm mark)

120 g * 10 cm = 50 g * ? cm
Answered by Jos kie kiez
To be answered
Answered by Wendy
Explain more please
Answered by Davy
Thumbs up
Answered by Vickie
Anyone are any of you real people
Answered by Vickie
Physics
A metre rule is balanced by masses of 24g and 16g suspended from its ends .find the position of the pivot??
Answered by Samson Opiyo
Yes
Answered by Samson Opiyo
help me solving it
Answered by Ibrahim
A uniform m metre rule of mass 110g is balance at xcm marks when a mass of 60g is lang at 10cm marks calculate. 1 the balance point x 2 the reaction at turning point
There are no AI answers yet. The ability to request AI answers is coming soon!