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zheka24 [161]
3 years ago
11

Please help, I don't get this

Physics
1 answer:
Trava [24]3 years ago
5 0
C the third one i think good luck
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Read 2 more answers
A​ DC-9 aircraft leaves an airport from a runway whose bearing is N4343degrees°E. After flying for one half 1 2 ​mile, the pilot
bearhunter [10]

Answer:

103.4° or S76.6°E

Explanation:

The direction N43°E is perpendicular to the direction south-east when the plane turn 90° and heads in the south-east direction.

Since the distance 1/2 mile N43°E is perpendicular to the distance 1 mile south-east, we have a right angled triangle.

So, the angle θ between the aircraft's new position and old position is gotten from tanθ = 1 ÷ 1/2 = 2

θ = tan⁻¹(2) = 63.43°

So, the total angle from North to its new position is 40° + 63.43° = 103.43°

Since we need the south-east bearing, the angle from south is 180° - 103.43° = 76.57° ≅ 76.6°

So, our bearing is 103.4° or S76.6°E

3 0
4 years ago
The Sears Tower in Chicago is approximately 444 m tall Suppose a book
Reptile [31]

Answer:

A. 66.0 m/s downwards

Explanation:

The Tower has a height of 444m

The book is dropped ,finding the velocity of the book 222m above the ground, means the book will be on air for a  height of 222 m .

Apply the formula for free fall in a horizontal projection as;

h= u²×sin²∅ /2g  where

h= maximum height =222m

g= acceleration due to gravity =9.81 m/s²

∅ = projectile angle = 0

u = velocity of the book

Applying the formula as ;

h= u²×sin²∅ /2g

222 = u²/2*9.81

222*19.62 = u²

4355.64 = u²

√4355.64 = u

65.99 m/s = u

66 m/s  downwards

4 0
3 years ago
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