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2 years ago

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You are in a hot-air balloon that, relative to the ground, has a veloc- ity of 6.0 m/s in a direction due east. You see a hawk m

oving directly away from the balloon in a direction due north. The speed of the hawk relative to you is 2.0 m/s. What are the magnitude and direction of the hawk’s velocity relative to the ground? Express the directional angle rel- ative to due east.

1 answer:

Ann [662]2 years ago

8 0

Answer:

6.32 m/s 18.43° northeast

Explanation:

We express the velocity of hawk as:

$v_{Hawk}=v_{balloon}+v_{HawkRelativetoBalloon}=6 x+2 y$

We consider positive x towards east and positive y due north. So the magnitude is simply the square root of the square components:

$|v_{hawk}|=\sqrt[]{6^2+2^2}=\sqrt{40}$ ≈ $6.32 m/s$

And the angle with respect to the east should be with:

$arctan(\frac{2}{6} )=18.43 \°$

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Answer:

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The moment of parallel pipe rotating about it's axis is given by the formula;

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Equation 4 becomes;

τ = $\frac{M}{3} (a^{2} +b^{2} )$ ∝

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I attached the missing picture.
We can figure this one out using the law of conservation of energy.
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1 year ago

Read 2 more answers

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We are given: Final velocity ( $v_f$ )=20 m/s .

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Subtracting 2.51a from both sides, we get

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Using another equation of motion

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Plugging values of vi =20-2.51a, t=2.51 and distnace s=82.9 in this equation.

We get,

$20-(20-2.51a)=2*a(82.90)$

Now, we need to solve it for a.

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a=0.

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Answer:

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