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3 years ago

Which substance is a combination of different atoms?

Physics

1 answer:

zvonat [6]3 years ago

6 0

Answer:

Heterogeneous

Explanation:

A chemical substance is composed of one type of atom or molecule. A mixture is composed of different types of atoms or molecules that are not chemically bonded. A heterogeneous mixture is a mixture of two or more chemical substances where the various components can be visually distinguished.

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Determine the ratio of the flow rate through capillary tubes A and B (that is, Qa/Qb).

Oliga [24]

To solve this problem we can use the concepts related to the change of flow of a fluid within a tube, which is without a rubuleous movement and therefore has a laminar fluid.

It is sometimes called Poiseuille’s law for laminar flow, or simply Poiseuille’s law.

The mathematical equation that expresses this concept is

$\dot{Q} = \frac{\pi r^4 (P_2-P_1)}{8\eta L}$

Where

P = Pressure at each point

r = Radius

$\eta =$ Viscosity

l = Length

Of all these variables we have so much that the change in pressure and viscosity remains constant so the ratio between the two flows would be

$\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}$

From the problem two terms are given

$R_A = \frac{R_B}{2}$

$L_A = 2L_B$

Replacing we have to

$\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}$

$\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_B^4}{16*r_B^4}\frac{L_B}{2*L_B}$

$\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{1}{32}$

Therefore the ratio of the flow rate through capillary tubes A and B is 1/32

6 0

3 years ago

PLEASE HELP ASAP 80 POINTS

weqwewe [10]

I'm sorry but I'm abouta fail a test if I dont d ig.i this for the points

4 0

3 years ago

Read 2 more answers

The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr

MA_775_DIABLO [31]

Answer:

$q=49Q/64$

and

$x =16R/15$

Explanation:

See attached figure.

$E_{Q}=$ E due to sphere

$E_{q}=$ E due to particule

$E_{total}=E_{Q}-E_{q}=0$ (1)

according to the law of gauss and superposition Law:

$E_{Q}=E_{1}+E_{2}=E_{2}$ ; electric field due to the small sphere with r1=R/4

$E_{Q}=kq_{2}/(r_{1}^{2})=$

$q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}$

then: $E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})$ (2)

on the other hand, for the particule:

$E_{q}=kq/(r_{p}^{2})$

$r_{p}=2R-R/4=7R/4$ ⇒ $E_{q}=16kq/(49R^{2})$ (3)

We replace (2) y (3) in (1):

$E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})$

$q=49Q/64$

--------------------

if R<x<2R AND $E_{total}=E_{Q}-E_{q}=0$

$E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})$

remember that $q=49Q/64$

then:

$Q(2R-x^{2})=49/64*x^{2}$

solving:

$x_{1} =16R/15$

$x_{2} =16R$

but: R<x<2R

so : $x =16R/15$

7 0

2 years ago

Goodmorning I can’t seem to remember this can someone help me

Bumek [7]

Explanation:

3 cause the triangle method of addition are connected to tips of one vector

6 0

2 years ago

an object of mass m is traveling at constant speed v in a circular path of radius r. how much work is done by the centripetal fo

vlada-n [284]

The work done is by the centripetal force on mass m during an angular displacement of 2π revolutions mv²2π /r J

Centripetal force - a force acts on an moving object in circular path.

the centripetal force is given by

F= mv²/r (equation1)

Work done is given by

W = Fd (equation 2)

d = 2π

work is done by the centripetal force on mass m during an angular displacement of 2π revolutions is given by:

to calculate work done using equation 1 in 2 we get

W = mv² d/r

W = mv² × 2π /r J

The work done is by the centripetal force on mass m during an angular displacement of 2π revolutions mv²2π /r J

To know more about centripetal force :

brainly.com/question/13031430

#SPJ4

6 0

1 year ago