Ask your own question, for FREE!
Physics 20 Online
OpenStudy (anonymous):

I dont get this :/ http://screencast.com/t/eSTW1oOF

OpenStudy (anonymous):

answer shuld b 10^2

OpenStudy (somy):

woah!! im confused here too T_T

OpenStudy (vincent-lyon.fr):

This is really food for thought!

OpenStudy (anonymous):

LOL awesome question \[\delta L = \frac{YLmg}{A}\] now.. all linear dimensions are reduced to one tenth L goes down by factor of 10 length of the cube goes down by factor of 10, (so u can calculate the new mass) radius of the wire goes down by a factor of 10 (so u can calculate the new area and so u can plug in and find the ratio of extensions!

OpenStudy (somy):

@Mashy

OpenStudy (anonymous):

ohhw :P.. no it should be 100

OpenStudy (somy):

why?

OpenStudy (somy):

marking scheme is showing B

OpenStudy (anonymous):

\[L ' = L/10 \\m' = m/1000 \\ A' = A/100\] so if u plug them in and find\[\delta L /\delta L'\] u would get a 100

OpenStudy (somy):

man =_= my question is different read it

OpenStudy (anonymous):

That is correct.. just plug in stress formula..

OpenStudy (somy):

how?

OpenStudy (somy):

i just divide all values by 10?

OpenStudy (anonymous):

all LINEAR.. so lenght and radius goes down by 10 but mass goes down by a 1000... think how

OpenStudy (somy):

by 1000????

OpenStudy (vincent-lyon.fr):

Mashy is right. The first answer is 100 The second answer is 10

OpenStudy (anonymous):

mass?

OpenStudy (anonymous):

@Mashy Thanks for your work, but we have only learnt the formula E =FL/AX where F = force, L = length , A = area X = extension, we havent learnt anythng about mass :/

OpenStudy (anonymous):

i dont understan, how u brought up mass:/

OpenStudy (vincent-lyon.fr):

The force exerted by the load is proportional to its mass, its volume, its length cubed. Hence real force is 10x10x10 times the model-force

OpenStudy (anonymous):

how did u gt tht equation>

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!