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4 years ago

What effect does the velocity of a rotating object have on the centripetal acceleration?

Physics

2 answers:

kipiarov [429]4 years ago

7 0

Answer:

as centripatal force acts upon an object moving at a circle in constant speed de force acts always inwards as de velocity of object is directed tangent to de circle de force can accelerate de object by changing its direction bt not actually de speed

alina1380 [7]4 years ago

5 0

The two quantities are closely related, but the cause/effect is the other way around.

-- The centripetal force is caused by something outside this discussion, not by the object.

-- The centripetal force acting on the object determines the object's centripetal acceleration.

-- The centripetal acceleration is the cause of whatever the object's velocity (speed and direction) turns out to be.

-- It's the centripetal acceleration that has the effect on the object's velocity.

As an example, you wouldn't say that the orbiting of a TV satellite is what causes the Earth's centripetal force that acts on it.

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3 years ago

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3 years ago

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2 years ago

Part 1

xz_007 [3.2K]

a) The x-coordinate of the center of gravity is $\frac{19}{18}\cdot a$ .

b) The y-coordinate of the center of gravity is $\frac{19}{18}\cdot a$ .

<h3>Determination of the coordinates of the center of gravity</h3>

Let suppose that each square has an uniform density, the coordinates of the center of gravity of each square with respect to the origin are, respectively:

$\vec r_{1} = (0.5\cdot a, 0.5\cdot a)$

$\vec r_{2} = (1.5\cdot a, 0.5\cdot a)$

$\vec r_{3} = (0.5\cdot a, 1.5\cdot a)$

$\vec r_{4} = (1.5\cdot a, 1.5\cdot a)$

The center of gravity of the entire system is found by applying definition of weighted averages:

$\vec r_{cg} = \frac{W_{1}\cdot \vec r_{1}+W_{2}\cdot \vec r_{2}+W_{3}\cdot \vec r_{3}+W_{4}\cdot \vec r_{4}}{W_{1}+W_{2}+W_{3}+W_{4}}$

Where $W_{1}$ , $W_{2}$ , $W_{3}$ and $W_{4}$ are weights of the each square, in newtons.

Now we proceed the coordinates of the center of gravity of the entire system:

$\vec r_{cg} = \frac{(50\,N)\cdot (0.5\cdot a, 0.5\cdot a) + (30\,N)\cdot (1.5\cdot a, 0.5\cdot a)+(30\,N)\cdot (0.5\cdot a, 1.5\cdot a)+(70\,N)\cdot (1.5\cdot a, 1.5\cdot a)}{50\,N+30\,N+30\,N+70\,N}$

$\vec r_{cg} = \frac{5}{18}\cdot (0.5\cdot a, 0.5\cdot a) +\frac{1}{6}\cdot (1.5\cdot a, 0.5\cdot a) +\frac{1}{6}\cdot (0.5\cdot a, 1.5\cdot a) + \frac{7}{18}\cdot (1.5\cdot a, 1.5\cdot a)$

$\vec r_{cg} = \left(\frac{19}{18}\cdot a, \frac{19}{18}\cdot a \right)$ $\blacksquare$

a) The x-coordinate of the center of gravity is $\frac{19}{18}\cdot a$ . $\blacksquare$

b) The y-coordinate of the center of gravity is $\frac{19}{18}\cdot a$ . $\blacksquare$

To learn more on center of gravity, we kindly invite to check this verified question: brainly.com/question/20662119

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2 years ago

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neonofarm [45]

Answer :It is a bar graph

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3 years ago