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3 years ago

Is smoke diffusion??

Physics

2 answers:

timofeeve [1]3 years ago

4 0

Yes, it is diffusion.

Ivanshal [37]3 years ago

3 0

Diffusion is the spread of particles through random motion from regions of high concentration to regions of lower concentration. Because diffusion works without barriers, when smoke is emitted into the air, it can easily travel to surrounding areas. hope this can help you ^_^

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Which type of mixture could this illustration represent?

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Answer:

A homogeneous mixture

Explanation:

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Metallurgy is the study of _____.

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Forming ionic bonds study

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Determine the CM of a rod assuming its linear mass density λ (its mass per unit length) varies linearly from λ = λ0 at the left

Dahasolnce [82]

Answer:

$x_c= \dfrac{5}{9}L$

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Explanation:

Here mass density of rod is varying so we have to use the concept of integration to find mass and location of center of mass.

At any distance x from point A mass density

$\lambda =\lambda_0+ \dfrac{2\lambda _o-\lambda _o}{L}x$

$\lambda =\lambda_0+ \dfrac{\lambda _o}{L}x$

Lets take element mass at distance x

dm =λ dx

mass moment of inertia

$dI=\lambda x^2dx$

So total moment of inertia

$I=\int_{0}^{L}\lambda x^2dx$

By putting the values

$I=\int_{0}^{L}\lambda_ ox+ \dfrac{\lambda _o}{L}x^3 dx$

By integrating above we can find that

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Now to find location of center mass

$x_c = \dfrac{\int xdm}{dm}$

$x_c = \dfrac{\int_{0}^{L} \lambda_ 0(1+\dfrac{x}{L})xdx}{\int_{0}^{L} \lambda_0(1+\dfrac{x}{L})}$

Now by integrating the above

$x_c=\dfrac{\dfrac{L^2}{2}+\dfrac{L^3}{3L}}{L+\dfrac{L^2}{2L}}$

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So mass moment of inertia $I=\dfrac {7}{12}\lambda_ 0 L^3$ and location of center of mass $x_c= \dfrac{5}{9}L$

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3 years ago

The moon's mass is 7.34x10-kg and it is 3.8x10m away from earth. Calculate the gravitational force of attraction between earth a

Nana76 [90]

Again I think you did not give the right constants. So I would use the correct constants for mass of moon and distance from earth to moon.

The formula for force of attraction between any two bodies in the universe
F = GMm / r^2. (Newton's Universal law of Gravitation).

G = Universal gravitational constant, G = 6.67 * 10 ^ -11 Nm^2 / kg^2.
M = Mass of Earth. = 5.97 x 10^24 kg.
m = mass of moon = 7.34 x 10^22 kg.
r = distance apart, between centers = in this case it is the distance from Earth to the Moon
= 3.8 x 10^8 m.

(Sorry I could not assume with the values you gave, they are wrong, and if we use them we would be insulting Physics).

So F = ((6.67 * 10 ^ -11)*(5.97 x 10^24)*(7.34 * 10^22)) / (3.8 x 10^8)^2.
Punch it all up in your calculator.

I used a Casio 991 calculator, it should be one of the best in the world.Really lovely calculator, that has helped me a lot in computations like this. I am thankful for the Calculator.

F = 2.0240 * 10^ 20 N.
So that's our answer.
Hurray!!

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Andre45 [30]

Answer:

50 %

Explanation:

It's Fifty percent

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