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3 years ago

12

If you push a book against a wall hard enough it will not slide down even though gravity is pulling it. Use what you know about

friction and newtons laws of motion to explain why the book does not fall

1 answer:

e-lub [12.9K]3 years ago

3 0

Yes the book will fall on the floor

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Compute the density in g/cm? of a piece of metal that has a mass of 0.450 kg and a volume of 52 cm3

Y_Kistochka [10]

Answer:

Ro = 8.65 [g/cm³]

Explanation:

We must remember that density is defined as the ratio of mass to volume.

$Ro=m/V$

where:

m = mass = 0.450 [kg] = 450 [g]

V = volumen = 52 [cm³]

Ro = density [g/cm³]

Now replacing:

$Ro = 450/52\\Ro = 8.65 [g/cm^{3} ]$

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2 years ago

Find the GCF 12, 18, and 84

ahrayia [7]

To find the answer, plot down the factors for every number.

12: 1, 2 ,3 ,4, 6, 12

18: 1, 2, 3, 6, 9, 18

84: 1, 2, 3, 4, 6, 7, 12

If you noticed, the number that was common to the 3 numbers, were 1, 2, 3, and 6
And 6 is the bigger number
So 6 is your GCF

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3 years ago

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After a match has burned which of these is not proof that a chemical reaction occurred?

mixas84 [53]

In my estimation I would say C, I was leaning towards A, but I believe that would merely be "incomplete combustion." I hope this was semi-helpful!

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Help on finding kinetic energy??

Jobisdone [24]

Trick question? In order to have kinetic energy, an object must be moving. Therefore, in this case, kinetic energy would be 0. If it were asking about potential energy, it would be a different story.

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A particle travels clockwise on a circular path of diameter R, monitored by a sensor on the circle at point P; the other endpo

kotykmax [81]

We make a graphic of this problem to define the angle.

The angle we can calculate through triangle relation, that is,

$sin\theta = \frac{c}{QP}\\sin\theta = \frac{c}{R}\\\theta=sin^{-1}\frac{c}{R}$

With this function we should only calculate the derivate in function of c

$\frac{d\theta}{dc} = \frac{1}{\sqrt{1-\frac{c^2}{R^2}}}(\frac{c}{R})'\\\frac{d\theta}{dc} = \frac{1}{\sqrt{R^2-c^2}}$

That is the rate of change of $\theta$ .

b) At this point we need only make a substitution of 0 for c in the equation previously found.

$\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{\sqrt{R^2-0}}\\\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{R}$

Hence we have finally the rate of change when c=0.

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