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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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Altitude refers to ________________. Group of answer choices the angular distance from the celestial equator the angular height
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2 years ago
A cat leaps into the air to catch a bird with an initial speed of 2.74 m/s at an angle of 60.0° above the ground. What is the hi
Volgvan

Answer: D. 0.29 m

Explanation:

We will use the following equations to describe the leap of the cat:

y=V_{o}sin\theta t-\frac{gt^{2}}{2}   (1)

V_{y}=V_{oy}-gt   (2)

Where:

y  is the height of the cat  

V_{oy}=V_{o}sin\theta is the cat's initial velocity

\theta=60\°

g=9.8m/s^{2}  is the acceleration due gravity

t is the time

V_{y} is the y-component of the velocity

Now the cat will have its maximum height y_{max} when V_{y}=0. So equation (2) is rewritten as:

0=V_{oy}-gt   (3)

Finding t:

t=\frac{V_{oy}}{g}=\frac{V_{o}sin\theta}{g}   (4)

t=\frac{2.74 m/s sin(60\°)}{9.8m/s^{2}}   (5)

t=0.24 s   (6)

Substituting (6) in (1):

y_{max}=(2.74 m/s)sin(60\°) (0.24 s)-\frac{(9.8m/s^{2})(0.24 s)^{2}}{2}   (7)

Finally:

y_{max}=0.287 m \approx 0.29 m   (8)

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A person jogs eight complete laps around a 400-m track in a total time of 14.5 min. Calculate the average speed and the average
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