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klasskru [66]
3 years ago
7

Find the magnitude of vector A = i - 2j + 3k O V14 10 O4

Physics
1 answer:
Angelina_Jolie [31]3 years ago
6 0

Answer:

|A|=\sqrt{1^2+(-2)^2+(3)^2}=3.74

Explanation:

Given that,

Vector A=i-2j+3k............(1)

We need to find the magnitude of vector A. Let us suppose vector in the form of, R = xi + yj + zk. Its magnitude is given by :

|R|=\sqrt{x^2+y^2+z^2}

In equation (1),

x = 1

y = -2

z = 3

So, the magnitude of vector A is given by :

|A|=\sqrt{1^2+(-2)^2+(3)^2}

|A| = 3.74

So, the magnitude of given vector is 3.74. Hence, this is the required solution.

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A vector A⃗ has a magnitude of 40.0 m and points in a direction 20.0∘ below the positive x axis. A second vector, B⃗, has a magn
jolli1 [7]

The magnitude of the vector C is 96.32m

<h3>How to solve for the magnitude of vector c</h3>

Ax = AcosθA

= 40 cOS 20

= 37.59

Ay = AsinθA

-40sin20

= -13.68

Bx = B cos θ B

= 75Cos50

= 48.21

By = BsinθB

= 75sin50

= 57.45

Cx = AX + Bx

= 37.59 + 48.21

= 85.8

Cy = Ay + By

= -13.65 + 57.45

= 43.77

The magnitude is solved by

|c| = \sqrt{Cx^{2}+Cy^{2}  }

= √85.8² + 43.77²

= 96.32m

The magnitude of the vector c is 96.32m

Read more on the magnitude of a vector here:

brainly.com/question/3184914

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Please help in desperate
Rom4ik [11]

Answer:

D. 53°

Explanation:

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What is the acceleration of a car that goes from 40 m/s to 80 m/s in 2s?
Law Incorporation [45]

Answer:

2

Explanation:

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