- #1

ago01

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- Homework Statement
- calculating the magnitude of the sum of vector A who has magnitude 13.0 and angle 38.7 degrees off the horizontal and a vector who has magnitude 5.0 and angle 26.0 degrees off the horizontal. Use proper sig figs.

- Relevant Equations
- ##|V| = sqrt(V_x^2 + V_y^2)##

In general I think I follow significant figures. On one and two step problems I don't really have such a problem but I seem to constantly miss points because I get lost in complicated ones. I'm hoping I can understand more as getting into science means getting friendly with these.

Let's suppose I am calculating the magnitude of the sum of vector A who has magnitude 13.0 and angle 38.7 degrees off the horizontal and a vector who has magnitude 5.0 and angle 26.0 degrees off the horizontal.

Then:

##R_x = (13.0)cos(38.7) + (5.0)cos(26.0)##

So the first product maintains 3 sig figs, and the second retains 2 sig figs.

##R_x = 7.014 + 4.49##

retaining one sig fig extra (my professor says to do this instead of holding the whole number).

##R_x = 11.508##

Retained to 1 extra sig fig (so now we have 2 decimals of precision available).

To the correct decimals of precision it's 11.51.

The same logic applies to #R_y#

##R_y = (13.0)sin(38.7) + (5.0)sin(26.0)##

##R_y = 10.94 + 3.81##

##R_y = 14.75##

In this case the extra retained SF in both the products ended up producing the right amount of degrees of precision here.

Then the magnitude of R

##|R| = \sqrt{11.508^2 + 14.75^2}##

Since the equations above say we really don't "have" the 8 in the first square we want 4 SF, and we don't really "have" the 5 in the second square so we want 3 SF there.

##|R| = \sqrt{132.43 + 217.7}##

Retaining one extra sig fig. So we have 1 decimal of precision available to us in the sum, so we will retain 2 to reduce rounding errors.

##|R| = \sqrt{350.13}##

and we have 4 sig figs here (since the 3 was carried over)

##|R| = 18.71##

But this can't be right.

Thank you!

Let's suppose I am calculating the magnitude of the sum of vector A who has magnitude 13.0 and angle 38.7 degrees off the horizontal and a vector who has magnitude 5.0 and angle 26.0 degrees off the horizontal.

Then:

##R_x = (13.0)cos(38.7) + (5.0)cos(26.0)##

So the first product maintains 3 sig figs, and the second retains 2 sig figs.

##R_x = 7.014 + 4.49##

retaining one sig fig extra (my professor says to do this instead of holding the whole number).

##R_x = 11.508##

Retained to 1 extra sig fig (so now we have 2 decimals of precision available).

To the correct decimals of precision it's 11.51.

The same logic applies to #R_y#

##R_y = (13.0)sin(38.7) + (5.0)sin(26.0)##

##R_y = 10.94 + 3.81##

##R_y = 14.75##

In this case the extra retained SF in both the products ended up producing the right amount of degrees of precision here.

Then the magnitude of R

##|R| = \sqrt{11.508^2 + 14.75^2}##

Since the equations above say we really don't "have" the 8 in the first square we want 4 SF, and we don't really "have" the 5 in the second square so we want 3 SF there.

##|R| = \sqrt{132.43 + 217.7}##

Retaining one extra sig fig. So we have 1 decimal of precision available to us in the sum, so we will retain 2 to reduce rounding errors.

##|R| = \sqrt{350.13}##

and we have 4 sig figs here (since the 3 was carried over)

##|R| = 18.71##

But this can't be right.

**The original question only had 2 SF available (5.0)!**So this should be 19 to our problem's precision no?I *think* I followed the logic correctly but I cannot figure out why I get*more*precision in the answer than I started with. That doesn't add up. Should I always round to the sig figs given to the problem or did I make another critical mistake here I am missing?Thank you!

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