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9 Group TitleBest ResponseYou've already chosen the best response.1
i need to see the steps of solving it plz :)
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.1
Do you know the relation between Pressure and force?
 one year ago

9 Group TitleBest ResponseYou've already chosen the best response.1
i know \[p = \frac{ F }{ a }\] So \[= \frac{ m a }{a }\]
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.1
\[P=ma / A\] \[a\ne A\] can you tell me the difference between a and A?
 one year ago

9 Group TitleBest ResponseYou've already chosen the best response.1
idk , u tell me :B
 one year ago

9 Group TitleBest ResponseYou've already chosen the best response.1
i just wanna know how we got L to the power 1 and T to the power 2
 one year ago

9 Group TitleBest ResponseYou've already chosen the best response.1
how we got the negative powers ?
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.1
a= acceleration= distance/ (time )^2= L/T^2=\(LT^{2}\) A= area= \(L^2\)
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.1
Can you try now?
 one year ago

9 Group TitleBest ResponseYou've already chosen the best response.1
= \[\frac{ M L T ^{2} }{L T ^{2} }\]
 one year ago

9 Group TitleBest ResponseYou've already chosen the best response.1
is that right 0
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.1
Why you have \(Lt^{2}\) in the denominator?
 one year ago

9 Group TitleBest ResponseYou've already chosen the best response.1
because pressure = \[\frac{ m a }{a }\] and \[a = l T ^{2}\]
 one year ago

9 Group TitleBest ResponseYou've already chosen the best response.1
u just said that >_>
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.1
Read my post again,
 one year ago

9 Group TitleBest ResponseYou've already chosen the best response.1
is p pressure Force over area ?
 one year ago

9 Group TitleBest ResponseYou've already chosen the best response.1
i thought its force over acceleration .
 one year ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.1
No problem, you got it now?
 one year ago

9 Group TitleBest ResponseYou've already chosen the best response.1
yeah , thanks again
 one year ago
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