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Andre45 [30]
4 years ago
9

A jet airliner moving initially at 889 mph

Physics
1 answer:
Eduardwww [97]4 years ago
8 0

Answer:

1500 mph

Explanation:

Take east to be +x and north to be +y.

The x component of the velocity is:

vₓ = 889 cos 0° + 830 cos 59°

vₓ = 1316.5 mph

The y component of the velocity is:

vᵧ = 889 sin 0° + 830 sin 59°

vᵧ = 711.4 mph

The speed is found with Pythagorean theorem:

v² = vₓ² + vᵧ²

v² = (1316.5 mph)² + (711.4 mph)²

v = 1496 mph

Rounded to two significant figures, the jet's speed relative to the ground is 1500 mph.

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(a) Differentiate the position vector to get the velocity vector:

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<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>

<em></em>

(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.

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A 2,300-kg truck is traveling down a highway at 32 m/s. What is the kinetic energy of the truck?
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m = mass of the truck = 23 00 kg

v = speed of the truck down the highway = 32 m/s

K = kinetic energy of the truck = ?

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K = (0.5) m v²

inserting the values

K = (0.5) (2300) (32)²

K = (0.5) (2300) (1024)

K = (1150) (1024)

K = 1177600 J

hence the kinetic energy of the truck comes out to be 1177600 J

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Answer:

The time is 1.8s

Explanation:

The ball droped, will freely fall under gravity.

Hence we use free fall formula to calculate the time by the ball to hit the ground

h= \frac{1}{2}g{t}^{2}

Where h is the height from which the ball is droped, g is the acceleration due to gravity that acted on the ball, and t is the time taken by the ball to hit the ground.

From the question,

h=16m

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By substitution we obtain,

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Diving through by 9.8

\frac{32}{9.8}= \frac{ 9.8{t}^{2} }{9.8}

\implies{t}^{2} =3.265

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