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Plz. explain how the uncertainty becomes angle using tan?


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Posted

tan(θ°) = (viv) / (vih)

= (1.28 ± 0.0676m/s) / (0.514 ± 0.0676m/s)

= 2.55 ± 0.467 m/s

θ° = 68.0 ± 3.65°

 

This answer may not be correct to the decimal place but it shud b relatively close. Could someone plz explain to me how 3.65° becomes the uncertainty for the angle? Plz. explain very clearly and simply. Im in a hurry.

 

THanks in advance

Posted
tan(?°) = (viv) / (vih)

= (1.28 ± 0.0676m/s) / (0.514 ± 0.0676m/s)

= 2.55 ± 0.467 m/s

?° = 68.0 ± 3.65°

 

This answer may not be correct to the decimal place but it shud b relatively close. Could someone plz explain to me how 3.65° becomes the uncertainty for the angle? Plz. explain very clearly and simply. Im in a hurry.

 

THanks in advance

First of all you shouldn't have the units that I've bolded.

 

You propagate errors by using the derivative:

 

d(tan x)/dx = 1/cos2x

 

so the uncertainty in tan x' date=' which is d(tan x), is the same as dx/cos[sup']2[/sup]x

 

d(tan x) = .467 from your work and x = 68 degrees, making dx=3.35 degrees

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