Aamc 11 #20

[typicalindian]

I feel pretty dumb for asking this but anyways...

why is the value for area 5.0x10^-4? Don't they tell you that the cross sectional area is 5.0 cm...which should be 5.0x10^-2m..no?

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[MrNeuro]

You need to convert to m^2 and cm^2 to m^2 is 10000 cm^2 in 1 m^2

 

[wanderedtoolong]

I got this one wrong too! Didn't even think about it- just glanced at the answer and figured I'd made a stupid calculation error. Now that you've pointed it out, yeah, what gives!?

 

[wanderedtoolong]

oooh now I see. . .

 

[typicalindian]

You need to convert to m^2 and cm^2 to m^2 is 10000 cm^2 in 1 m^2

ohhhh ok thanks

 

[pfaction]

I had no idea as I also got this wrong. I saw cm^-2 but didn't factor it in, ****s.

 

[MedPR]

Always had a hard time with this. It helps to write it out a few times until you get used to it.

(5cm^2)*(10^-2m/cm) = 5*10-2cm*m

(5*10^-2cm*m)*(10^-2m/cm) = 5*10^-4m^2

 

[BrianK0220]

How does delta L divided by L = 0.01? I don't see how that's equal to a one percent change in length. Wouldn't delta L alone equal 0.01?

 

[brownbaglunch]

How does delta L divided by L = 0.01? I don't see how that's equal to a one percent change in length. Wouldn't delta L alone equal 0.01?

Delta L is one percent of the original length:

 

[BrianK0220]

Perfect, God bless you!