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A ship moves through the water at 30 miles/hour at an angle of 30° south of east. The water is moving 5 miles/hour at an angle of 20° east of north. Identify the ship's vector, the water current's vector, and the vector representing the ship's actual motion.

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Аноним Аноним · Answer 1 · 2019-07-25T13:21:03+02:00

Answer:

See below in bold.

Step-by-step explanation:

Ship's vector:

Horizontal component = 30 cos 30 = 25.98.

Vertical component = 30 sin(-30) = -15.

So it is <25.98, -15).

The current's vector:

Horizontal component = 5 sin 20 = 1.71.

Vertical component = 5 cos 20 = 4.7.

So it is <1.71, 4.7>.

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nuuk nuuk · Answer 2 · 2019-08-14T09:18:19+02:00

Answer:

Step-by-step explanation:

Resultant R of the two vectors is

[tex]R^2=a^2+b^2+2abcos\theta [/tex]

[tex]R^2=30^2+5^2+2\left ( 30\right )\left ( 5\right )cos\theta [/tex]

Where [tex]\theta [/tex]is the angle between Two vectors which is 100^{\circ}[/tex]

[tex]R^2=900+25+2\times 30\times 5\times cos100[/tex]

[tex]R^2=925-52.094[/tex]

R=29.544 miles per hour

and it makes an angle [tex]\phi [/tex]with 30 miles/hr vector

[tex]tan\phi =\frac{5}{30}[/tex]

[tex]\phi =tan^{-1}\frac{1}{6}[/tex]

[tex]\phi =9.462^{\circ}[/tex]

i.e.it makes an angle of 20.538 South of east with horizontal