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dybincka [34]
3 years ago
5

Bryan Allen pedaled a human-powered aircraft across the English Channel from the cliffs of Dover to Cap Gris-Nez on June 12, 197

9. (a) He flew for 169 min at an average velocity of 3.53 m/s in a direction 45° south of east. What was his total displacement? (Enter the magnitude in km and the direction in degrees south of east.) magnitude km direction ° south of east (b) Allen encountered a headwind averaging 2.00 m/s almost precisely in the opposite direction of his motion relative to the Earth. What was his average velocity relative to the air? (Enter the magnitude in m/s and the direction in degrees south of east.) magnitude m/s direction ° south of east (c) What was his total displacement relative to the air mass? (Enter the magnitude in km and the direction in degrees south of east.) magnitude km direction ° south of east
Physics
1 answer:
Leno4ka [110]3 years ago
3 0

(a) 3.58 km 45° south of east

The total displacement is given by:

d=vt

where

v is the average velocity

t is the time

The average velocity is:

v = 3.53 m/s

While we need to convert the time from minutes to seconds:

t=169 min \cdot 60 s/min = 10140 s

Therefore, the magnitude of the displacement is

d=(3.53)(10140)=35794 m = 3.58 km

And the direction is the same as the velocity, therefore 45° south of east.

(b) 5.53 m/s 90° south of east

The velocity of the air relative to the ground is

v_a = 2.00 m/s

and the direction is exactly opposite to that of Allen, so it is 45° north of west. Allen's velocity relative to the ground is

v = 3.53 m/s

So this must be the resultant of Allen's velocity relative to the air (v') and the air's speed (v_a). Since these two vectors are in opposite direction, we have

v= v'-v_a

Therefore we find v', Allen's velocity relative to the air:

v'=v+v_a = 3.53 + 2.00 = 5.53 m/s

The direction must be measured relative to the air's reference frame. In this reference frame, Allen is moving exactly backward, so his direction will be 90° south of east.

(c) 56.1 km at 90° south of east.

Since Allen's velocity relative to the air is

v' = 5.53 m/s

Then the displacement of Allen relative to the air will be given by

d'=v't

and substituting,

d'=(5.53)(10140)=56074 m = 56.1 km

And the direction is the same as that of the velocity, therefore will be 90° south of east.

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Answer:

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A₂ = \frac{\pi}{4}0.05^2 = 1.96\times 10^{-3}m^2

g = acceleration due to gravity = 9.8m/s²

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The gauge pressure at throat = Absolute pressure - The atmospheric pressure

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Substituting the values in the discharge formula we get

Q=0.97\frac{7.85\times 10^{-3}\times 1.96\times 10^{-3}}{\sqrt{7.85\times 10^{-3}^2-1.96\times 10^{-3}^2}}\sqrt{2\times 9.8\times 11.3}

or

Q=\frac{0.97\times15.42\times 10^{-6}\times 14.88}{7.605\times 10^{-3}}

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Answer:

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