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tiny-mole [99]
3 years ago
7

A light plane must reach a speed of 33 m/s per take off. How long a runway is needed if the constant acceleration is 3.0 m/s^2.

Physics
1 answer:
aleksley [76]3 years ago
3 0
Given the final velocity (Vf) and the acceleration (a), the distance that should be traveled by the plane is calculated through the equation,
                            d = (Vf² - Vi²) / 2a
V1 should be zero because the light plane started the motion from rest. Substituting the given values,
                          d = ((33 m/s)² - 0)) / 2(3 m/s²)
The distance is therefore equal to 181.5 meters. 
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Answer:

Value of angle between vector a and b is 56.30^{\circ}.

Explanation:

Vectors a and b have scalar product 6.00

Let \theta be the angle between a and b.

\vec{a}.\vec{b} = 6

ab cos \theta = 6 ...(1)

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\vec{a} \times\vec{b} = 9

ab sin \theta = 9 ...(2)

Dividing equation (2) by (1) we get

\frac{ab sin \theta}{ab cos \theta}  = \frac{9}{6}

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Explanation:

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