Can You All Tell Me How To Do This Step By Step.

  • Thread starter Thread starter Chikawakajones
  • Start date Start date
AI Thread Summary
The discussion revolves around a geometry problem involving two adjacent squares with areas of 4 cm² and 196 cm². The first square has a side length of 2 cm and the second has a side length of 14 cm, leading to inscribed circles with radii of 1 cm and 7 cm, respectively. Participants calculate the distance between the centers of the circles as 8 cm. There is some confusion regarding the term "adjacent" and the lack of a visual representation of the squares. The conversation highlights the importance of clear communication in geometry problems.
Chikawakajones
Messages
22
Reaction score
0
:confused: The Areas Of 2 Adjacent Squares Are 4 cm^2 and 196 cm^2.

Find The Length Of The Segment Joining The Centers Of Their Inscribe Circles.

(Theres A Picture Of 1 Big Square And 1 Small Square - Each Square Has A Circle Inside - The Center Of Each Circle Is Indicated By A Dot - There Is One Line Connecting The 2 Dots.)

( I Don't Know How To Post The Picture In Here)
 
Physics news on Phys.org
well, the first square is 2x2 (you just get the square root) and the 2nd square is 14x14 so...
the radius of the first circle is 1 (it's half the diameter) and the 2nd one is 7 so...?

7+1 = 8
 
futb0l said:
well, the first square is 2x2 (you just get the square root) and the 2nd square is 14x14 so...
the radius of the first circle is 1 (it's half the diameter) and the 2nd one is 7 so...?

7+1 = 8

Since this is a geometry problem without a picture and with an ambiguous text (u cannot tell only by looking at the text how those squares are),text it is quite a big probability that uyour answer to be wrong. :wink:

Daniel.

PS.Did u try to put as an attachement to your message?It usually works that way,as long as it's not more than 50kB of data.
 
dextercioby said:
Since this is a geometry problem without a picture and with an ambiguous text (u cannot tell only by looking at the text how those squares are),text it is quite a big probability that uyour answer to be wrong. :wink:

Daniel.

PS.Did u try to put as an attachement to your message?It usually works that way,as long as it's not more than 50kB of data.

You're right Dan - My name is also Daniel too, btw :-p ...
I don't know what he meant by "adjacent" - I just assume the simplest case.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top