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Jet001 [13]
3 years ago
11

A person on a cruise ship is doing laps on the promenade deck. On one portion of the track the person is moving north with a spe

ed of 3.6 m/s relative to the ship. The ship moves east with a speed of 12 m/s relative to the water. Part A What is the direction of motion of the person relative to the water?
Physics
1 answer:
Aleksandr-060686 [28]3 years ago
7 0

T<u>he direction of motion</u> of the person relative to the water is <u>16.7° north of east.</u>

Why?

We can solve the problem by applying the Pitagorean Theorem, where the first speed (to the north) and the second speed (to the east) corresponds to two legs of the right triangle formed with them. (north and east directions are perpendicular each other)

We can calculate the angle that give the direction using the following formula:

Tan(\alpha)=\frac{NorthSpeed}{EastSpeed}\\\\Tan(\alpha)^{-1}=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}\\\\\alpha=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}

Now, substituting the given information we have:

alpha=Tan(\frac{NorthSpeed}{EastSpeed})^{-1}

\alpha =Tan(\frac{3.6\frac{m}{s} }{12\frac{m}{s} })^{-1}\\\\\alpha =Tan(0.3)^{-1}=16.69\°(North-East)=16.7\°(North-East)

Hence, we have that <u>the direction of motion</u> of the person relative to the water is 16.7° north of east.

Have a nice day!

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We can rewrite (1) as

F_m = \frac{F_f}{23.4}

And inserting this into (2):

F=F_f + \frac{F_f}{23.4}

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F=F_f + F_m (2)

Using

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A=340 mm^2=340\cdot 10^{-6}m^2 is the total cross-sectional area

Substituting V_f=0.3, we have

A_f = (340\cdot 10^{-6} m^2)(0.3)=102\cdot 10^{-6} m^2

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The cross-sectional area of the matrix phase is

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A=340 mm^2=340\cdot 10^{-6}m^2 is the total cross-sectional area

Substituting V_m=0.7, we have

A_m = (340\cdot 10^{-6} m^2)(0.7)=238\cdot 10^{-6} m^2

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So, the strain experienced by the composite is

\epsilon=\frac{\sigma}{E}=\frac{0.137 GPa}{41.0 GPa}=3.34\cdot 10^{-3}

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