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3 years ago

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3. One atom of silicon can properly be combined in a compound with A. One atom of chlorine. B. Three atoms of hydrogen. C. Two a

toms of oxygen. D. Four atoms of calcium.

1 answer:

Paraphin [41]3 years ago

4 0

C :<em>2 atoms of Oxygen can probably join 1 Silicon Atom</em>

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A yo-yo is made from two uniform disks, each with mass m and radius R, connected by a light axle of radius b. A light, thin stri

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Answer:

linear acceleration

$a = \frac{2g}{2 + \frac{R}{b}}$

angular acceleration

$\alpha = \frac{2g}{R(2 + \frac{R}{b})}$

Explanation:

As we know that the force due to tension force is upwards while weight of the disc is downwards

so we will have

$2mg - T = 2ma$

also we have

$Tb = (\frac{1}{2}mR^2 + \frac{1}{2}mR^2)\alpha$

now we have

$Tb = mR^2(\frac{a}{R})$

$T = \frac{mRa}{b}$

now we have

$2mg = (2ma + \frac{mRa}{b})$

$a(2 + \frac{R}{b}) = 2g$

so we have

linear acceleration

$a = \frac{2g}{2 + \frac{R}{b}}$

angular acceleration

$\alpha = \frac{2g}{R(2 + \frac{R}{b})}$

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3 years ago

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$sound \: wave \: \\ light \: wave$

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2 years ago

A stone is tied to a string and whirled at constant speed in a horizontal circle. The speed is then doubled without changing the

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Answer:

Afterward the magnitude of the acceleration of the stone is four times as great

Explanation:

A stone tied to a string and whirled at a constant speed in a horizontal circle will have a centripetal acceleration given by

$a = \frac{v^{2} }{r}$

Where $a$ is the centripetal acceleration

$v$ is the speed

and $r$ is the radius of the circle

Here, the radius of the circle is the length of the string.

Now, from the question

The speed is then doubled without changing the length of the string,

Let the new speed be $v_{2}$ , that is

$v_{2} = 2v$

and without changing the length of the string means radius $r$ is constant.

To determine the magnitude of the acceleration of the stone afterwards,

Let the new acceleration be $a_{2}$ .

Then we can write that

$a_{2} = \frac{v_{2}^{2} }{r}$

From

$a = \frac{v^{2} }{r}$

$v = \sqrt{ar}$

Recall that

$v_{2} = 2v$

∴ $v_{2} = 2\sqrt{ar}$

Now, we will put the value of $v_{2}$ into

$a_{2} = \frac{v_{2}^{2} }{r}$

Then,

$a_{2} = \frac{(2\sqrt{ar}) ^{2} }{r}$

$a_{2} = \frac{4ar }{r}$

$a_{2} = 4a$

The new magnitude of the acceleration of the stone is four times the initial acceleration.

Hence,

Afterward the magnitude of the acceleration of the stone is four times as great

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