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3 years ago

Can anyone help me with this problem? It is -2.1??

Physics

1 answer:

nata0808 [166]3 years ago

8 0

Question not clear or read-able

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2 years ago

Explain why ultra-high voltages are<br> used to carry electricity over transmission lines.

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3 years ago

Read 2 more answers

The three vectors in Fig. 3-33 have magnitudes a = 3.00 m, b = 4.00 m, and c = 10.0 m and angle θ = 30.0°. What are

aliya0001 [1]

Answer:

a) $a_{x} =3$ b) $a_{y} =0$

c) $b_{x} =3.46$ d) $b_{y} =2$

e) $c_{x} =0$ f) $c_{y} =10$

g) p = -5.77 h) q=5

Explanation:

Diagram for given question is attached below in fig 1

for vector $\vec{a}$

θ = 0°

$a_{x} = 3 cos (0)\\a_{x} = 3\\a_{y} = 3 sin (0)\\a_{y} = 0$

for vector $\vec{b}$

θ = 30°

$b_{x} = 4 cos (30)\\b_{x} = 3.46\\b_{y} = 4 sin (30)\\b_{y} = 2$

for vector $\vec{c}$

θ = 90°

$c_{x} = 10 cos (90)\\c_{x} = 0\\c_{y} = 10 sin (90)\\c_{y} = 10$

$\vec{c} = p\vec{a} + q\vec{b}$

$c =c_{x} \hat{i} + c_{y}\hat{j}\\as a_{y} =0\\c_{x} \hat{i} + c_{y}\hat{j} = pa_{x} \hat{i} +q(b_{x} \hat{i} +b_{y} \hat{j} )\\c_{x} \hat{i} = pa_{x} \hat{i} +qb_{x} \hat{i}\\c_{y}\hat{j} =qb_{y}\hat{j}$

$q=\frac{c_{y}}{b_{y}} \\q=\frac{10}{2}\\q=5$

$c_{x} \hat{i} = pa_{x} \hat{i} +qb_{x} \hat{i}\\0 = p(3) + (5)(3.46)\\p = -5.77$

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