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Aloiza [94]
3 years ago
6

A slanted vector has a magnitude of 41 N and is at an angle of 23 degrees north of east. What are the magnitude and direction of

the horizontal and vertical components of this vector?
Physics
1 answer:
jenyasd209 [6]3 years ago
7 0

Answer:

The horizontal component is 37.74 N to east

The vertical component is 16.02 N to North

Explanation:

Any slant vector has two components

→ Horizontal component = R cos Ф

→ Vertical component = R sin Ф

→ R is the magnitude of the vector

→ Ф is the direction of the vector with positive part of the horizontal axis

A slanted vector has a magnitude of 41 N and is at an angle of 23

degrees north of east

23° north of east means the angle between the vector and the

east direction is 23° (east is the positive horizontal direction)

That means R is 41 N and Ф is 23°

→ R = 41 N , Ф = 23°

→ The horizontal component = 41 × cos(23) = 37.74 N east

→ The vertical component = 41 × sin(23) = 16.02 N North

<em>The horizontal component is 37.74 N to east</em>

<em>The vertical component is 16.02 N to North</em>

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<h3>What is a hypothesis?</h3>

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1 year ago
A satellite in outer space is moving at a constant velocity of 20.5 m/s in the +y direction when one of its on board thruster tu
Katyanochek1 [597]

Answer:

a)  V_f = 25.514 m/s

b)  Q =53.46 degrees CCW from + x-axis

Explanation:

Given:

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- Acceleration a = 0.31 i m/s^2

- Time duration for acceleration t = 49.0 s

Find:

(a) What is the magnitude of the satellite's velocity when the thruster turns off?

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Solution:

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                                   V_f = V_i + a*t

                                   V_f = 20.5 j + 0.31 i *49

                                   V_f = 20.5 j + 15.19 i

- The magnitude of the velocity vector is given by:

                                   V_f = sqrt ( 20.5^2 + 15.19^2)

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- The direction of the velocity vector can be computed by using x and y components of velocity found above:

                                 tan(Q) = (V_y / V_x)

                                 Q = arctan (20.5 / 15.19)

                                 Q =53.46 degrees

- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.

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