Question #b8942

Question

Question #b8942

1 Answer

Impact of this question

1237 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-13T02:38:36

Given
$\to X + \to Y = \to R$ $vec(X)+vec(Y)=vec(R)$

So $\to X + \to Y + \to Z = \to R + \to Z$ $vec(X)+vec(Y)+vec(Z)=vec(R)+vec(Z)$

Ler the angle between $\to Z and \to R$ $vec(Z) and vec(R)$ be $θ = 170^{\circ}$ $theta=170^@$

Also given $∣ ∣ ∣ \to Z ∣ ∣ ∣ = 13.5 N and ∣ ∣ ∣ \to R ∣ ∣ ∣ = 19.3 N$ $abs(vec(Z))=13.5N and abs(vec(R))=19.3N$

Now $∣ ∣ ∣ \to X + \to Y + \to Z ∣ ∣ ∣ = ∣ ∣ ∣ \to Z + \to R ∣ ∣ ∣$ $abs(vec(X)+vec(Y)+vec(Z))= abs(vec(Z)+vec(R))$

$= \sqrt{{∣ ∣ ∣ \to Z ∣ ∣ ∣}^{2} + {∣ ∣ ∣ \to R ∣ ∣ ∣}^{2} + 2 ∣ ∣ ∣ \to Z ∣ ∣ ∣ \cdot ∣ ∣ ∣ \to R ∣ ∣ ∣ cos θ}$ $=sqrt(abs(vec(Z))^2+ abs(vec(R))^2+2abs(vec(Z))*abs(vec(R))costheta)$

$= \sqrt{{(13.5)}^{2} + {(19.3)}^{2} + 2 \cdot 13.5 \cdot (19.3) \cdot cos 170}$ $=sqrt((13.5)^2+ (19.3)^2+2*13.5*(19.3)*cos170)$
$= 6.45 N$ $=6.45N$

Science

Math

Humanities

... and beyond

Question #b8942

1 Answer

Related questions

Impact of this question