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12 N in the opposite direction, forming an gle of 30 degrees with the positive x axis....what was the original question that you guys were discussing?
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12 N in the opposite direction, forming an gle of 30 degrees with the positive x axis....what was the original question that you guys were discussing?
How 12 and 30 ?12 N in the opposite direction, forming an gle of 30 degrees with the positive x axis....what was the original question that you guys were discussing?
Oh sorry what question? I kinda skipped over most of these comments cause I didn't have the time
Apply lamis ruleGuys..how the hell do we solve this question
9709/41/ON/10
q3??
i cant e ven resolve the forces
pm me if u can!!
The three forces are initially inequilibrium ie the resultant is zero. If u remove one of the forces the resultant will then be of the same magnitude as the removed force but in the exact opposite directionHow 12 and 30 ?
What year?
Apply lamis rule
What year?
You have a triangle p1XB...you can calculate P1X cause its the tentionib that string and XB is the tention in that respective string...so appy the cos rule...sorry I don't have the question in front of me so I might be confusing the labels. Anyway you can calculate the third angle by subtracting the calculated one frpm 90how are we going to find the angles
because that is the force which is bringing equilibrium to the system.If it is released or removed the resultant will be completely opposite in direction which is 30 degrees clockwise from +ve x-axis.How 12 and 30 ?
You have a triangle p1XB...you can calculate P1X cause its the tentionib that string and XB is the tention in that respective string...so appy the cos rule...sorry I don't have the question in front of me so I might be confusing the labels. Anyway you can calculate the third angle by subtracting the calculated one frpm 90
You have a triangle p1XB...you can calculate P1X cause its the tentionib that string and XB is the tention in that respective string...so appy the cos rule...sorry I don't have the question in front of me so I might be confusing the labels. Anyway you can calculate the third angle by subtracting the calculated one frpm 90
you understood what she explained??.:/..thanks
Meko bhi samjhado ji :/you understood what she explained??.:/..
Can u explain the workings??
PM me!!
you understood what she explained??.:/..
Can u explain the workings??
PM me!!
You have a triangle p1XB...you can calculate P1X cause its the tentionib that string and XB is the tention in that respective string...so appy the cos rule...sorry I don't have the question in front of me so I might be confusing the labels. Anyway you can calculate the third angle by subtracting the calculated one frpm 90
Yeah. And p1X is the weight of Ais XB equal to the weight of B?
I do keep SI units in my P1 Idk about P4 :/If the distance is given in cm...do we convert it into meters? Are the equations only applicable when the quantities are in their SI units?
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