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lara31 [8.8K]
3 years ago
14

The velocity of the wind relative to the water is crucial to sailboats. Suppose a sailboat is in an ocean current that has a vel

ocity of 2.20 m/s in a direction 30.0º east of north relative to the Earth. It encounters a wind that has a velocity of 4.50 m/s in a direction of 50.0º south of west relative to the Earth. What is the velocity of the wind relative to the water?
Physics
1 answer:
Brilliant_brown [7]3 years ago
6 0

Answer:

Explanation:

We shall write the velocities given in vector form to make the solution easy.

The velocity of water with respect to earth that is waV(e) makes 30 degree with north or 60 degree with east so in vector form

waV(e) = 2.2 cos 60 i + 2.2 sin 60 j

waV(e) = 1.1 i + 1.9 j

Similarly , velocity of wind with respect to earth that is wiV(e) , is making 50 degree with west or - ve of x axes so we cal write it in vector form as follows

wiV(e) =  - 4.5 cos 50 i - 4.5 sin 50 j

wiV(e) = - 2.89 i - 3.45 j

Now we have to calculate velocity of wind with respect to water that is

wiVwa

wiV( wa) = wiV ( e)+ eV(wa)

=  wiV( e)- waV(e)

- 2.89 i - 3.45 j - 1.1 i -  1.9 j

= - 3.99 i - 5.35 j

Magnitude of this relative velocity

D² = 3.99² + 5.35²

d = 6.67 m /s

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0.5-lbm of a saturated vapor is converted to asaturated liquid by being cooled in a weighted piston-cylinder device maintained a
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Answer:

The boiling point temperature of this substance when its pressure is 60 psia is  480.275 R

Explanation:

Given the data in the question;

Using the Clapeyron equation

(\frac{dP}{dT} )_{sat } = \frac{h_{fg}}{Tv_{fg}}

(\frac{dP}{dT} )_{sat } = \frac{\frac{H_{fg}}{m} }{T\frac{V_{fg}}{m} }

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(\frac{dP}{dT} )_{sat } =( \frac{250Btu\frac{778Ibf-ft}{Btu} }{0.5}) / ( (15+460)\frac{1.5}{0.5})  

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we substitute;

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