How many oxygen (O) atoms are in a molecule of Cg H,O3?<br> A. 1<br> B. 3<br> C. 10<br> D. 4

A person walks the path shown below. The total trip consists of four straight-line paths.

dmitriy555 [2]

At the end of the walk, the person's resultant displacement is 495.1 m at 63⁰ south of west.

<h3>What is resultant displacement?</h3>

The resultant displacement of an object is the change in position of the object. It can be described as the shortest distance connecting the final position of the object to the initial position of the object.

<h3>Net horizontal displacement </h3>

Path 1 = 40 m

Path 2 = 0 m

Path 3 = 110 m x cos(30) = 95.26 m

Path 4 = 180 m x cos(60) = 90 m

Total horizontal displacement, X = 40 m + 0 m + 95.26 m + 90 m = 225.26 m

<h3>Net vertical displacement </h3>

Path 1 = 0 m

Path 2 = 230 m

Path 3 = 110 m x sin(30) = 55 m

Path 4 = 180 m x sin(60) = 155.885 m

Total horizontal displacement, Y = 0 m + 230 m + 55 m + 155.885 m = 440.885 m

<h3>Resultant displacement</h3>

R = √(X² + Y²)

R = √(225.26² + 440.885²)

R = 495.1 m

<h3>Direction of the displacement</h3>

θ = arc tan (Y/X)

θ = arc tan (440.885/225.26)

θ = 63⁰

Thus, at the end of the walk, the person's resultant displacement is 495.1 m at 63⁰ south of west.

Learn more about resultant displacement here: brainly.com/question/13309193

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8 months ago

In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be

LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

The first mass is $m_1 = 0.42kg$

The second mass is $m_2 = 0.47kg$

From the question we can see that at equilibrium the moment about the point where the string holding the bar (where $m_1 \ and \ m_2$ are hanged ) is attached is zero

Therefore we can say that

$m_1 * 15cm = m_2 * xcm$

Making x the subject of the formula

$x = \frac{m_1 * 15}{m_2}$

$= \frac{0.42 * 15}{0.47}$

$x = 13.4 cm$

Looking at the diagram we can see that the tension T on the string holding the bar where $m_1 \ and \ m_2$ are hanged is as a result of the masses ( $m_1 + m_2$ )