They are important because numerous serve as areas where water vapour can condense.
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</h3><h3>What is condensation?</h3>
The transformation of water vapour into liquid is known as condensation. The process is the opposite of evaporation, in which liquid water turns into a vapour. Either the air is chilled to its dew point or it gets too saturated with water vapour to retain any more water, causing condensation to occur.
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Answer:
The answer to your question is: C. -9.81 m/s²
Explanation:
A. 9.81 m/s² acceleration is considered positive when it goes to the center of the earth, so this option is incorrect.
B. 0 m/s² This option is incorrect because acceleration is 0 for a linear motion without acceleration.
C. -9.81 m/s² If a projectile goes to the sky, then the acceleration will be negative.
D. It is not constant. Acceleration is constant.
(a) Differentiate the position vector to get the velocity vector:
<em>r</em><em>(t)</em> = (3.00 m/s) <em>t</em> <em>i</em> - (4.00 m/s²) <em>t</em>² <em>j</em> + (2.00 m) <em>k</em>
<em>v</em><em>(t)</em> = d<em>r</em>/d<em>t</em> = (3.00 m/s) <em>i</em> - (8.00 m/s²) <em>t</em> <em>j</em>
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(b) The velocity at <em>t</em> = 2.00 s is
<em>v</em> (2.00 s) = (3.00 m/s) <em>i</em> - (16.0 m/s) <em>j</em>
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(c) Compute the electron's position at <em>t</em> = 2.00 s:
<em>r</em> (2.00 s) = (6.00 m) <em>i</em> - (16.0 m) <em>j</em> + (2.00 m) <em>k</em>
The electron's distance from the origin at <em>t</em> = 2.00 is the magnitude of this vector:
||<em>r</em> (2.00 s)|| = √((6.00 m)² + (-16.0 m)² + (2.00 m)²) = 2 √74 m ≈ 17.2 m
(d) In the <em>x</em>-<em>y</em> plane, the velocity vector at <em>t</em> = 2.00 s makes an angle <em>θ</em> with the positive <em>x</em>-axis such that
tan(<em>θ</em>) = (-16.0 m/s) / (3.00 m/s) ==> <em>θ</em> ≈ -79.4º
or an angle of about 360º + <em>θ</em> ≈ 281º in the counter-clockwise direction.
