A community for students.
Here's the question you clicked on:
 0 viewing

This Question is Closed

9
 2 years ago
Best ResponseYou've already chosen the best response.1i need to see the steps of solving it plz :)

ash2326
 2 years ago
Best ResponseYou've already chosen the best response.1Do you know the relation between Pressure and force?

9
 2 years ago
Best ResponseYou've already chosen the best response.1i know \[p = \frac{ F }{ a }\] So \[= \frac{ m a }{a }\]

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

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

9
 2 years ago
Best ResponseYou've already chosen the best response.1how we got the negative powers ?

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

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

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

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

9
 2 years ago
Best ResponseYou've already chosen the best response.1is p pressure Force over area ?

9
 2 years ago
Best ResponseYou've already chosen the best response.1i thought its force over acceleration .

ash2326
 2 years ago
Best ResponseYou've already chosen the best response.1No problem, you got it now?
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.