Se arrastra un bloque de 18 kg a través de una superficie con una fuerza de 300 N en donde el coeficiente de fricción de dicha s
1 answer:
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Answer:
Solution ( Second Attachment ) : - 2.017 + 0.656i
Solution ( First Attachment ) : 16.140 - 5.244i
Step-by-step explanation:
Second Attachment : The quotient of the two expressions would be the following,
÷
So if we want to determine this expression in standard complex form, we can first convert it into trigonometric form, then apply trivial identities. Either that, or we can straight away apply the following identities and substitute,
( 1 ) cos(x) = sin(π / 2 - x)
( 2 ) sin(x) = cos(π / 2 - x)
If cos(x) = sin(π / 2 - x), then cos(2π / 5) = sin(π / 2 - 2π / 5) = sin(π / 10). Respectively sin(2π / 5) = cos(π / 2 - 2π / 5) = cos(π / 10). Let's simplify sin(π / 10) and cos(π / 10) with two more identities,
( 1 )
( 2 )
These two identities makes sin(π / 10) = , and cos(π / 10) =
.
Therefore cos(2π / 5) = , and sin(2π / 5) =
. Substitute,
÷
Remember that cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting those values,
÷
And now simplify this expression to receive our answer,
÷
=
,
=
and
=
=
As you can see our solution is option c. - 2.01749 was rounded to - 2.017, and 0.65552 was rounded to 0.656.
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First Attachment : We know from the previous problem that cos(2π / 5) = , sin(2π / 5) =
, cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting we receive a simplified expression,
We know that and
. Therefore,
Solution :
Which rounds to about option b.