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2 years ago

8

Determine the ratio of the flow rate through capillary tubes A and B (that is, QA/QB). The length of A is twice that of B, and t

he radius of A is one-half that of B. The pressure across both tubes is the same. Express your answer using three significant figures.

1 answer:

Jet001 [13]2 years ago

7 0

Answer:

$\dfrac{Q_A}{Q_B}=0.031$

Explanation:

Lets take

Radius of tube A=r

Length of tube A=L

Radius of tube B= r'

Length of tube B=L'

Given that

L = 2 L'

r= 0.5 r'

r' = 2 r

The pressure across tube given as

$\Delta p=\dfrac{8\mu LQ}{\pi R^{4}}$

$\dfrac{L_AQ_A}{ R_A^{4}}=\dfrac{L_BQ_B}{ R_B^{4}}$

$\dfrac{Q_A}{Q_B}=\dfrac{R_A^4}{R_B^4}\times \dfrac{L_B}{L_A}$

$\dfrac{Q_A}{Q_B}=\dfrac{r^4}{(2r)^4}\times \dfrac{L'}{2L'}$

$\dfrac{Q_A}{Q_B}=\dfrac{1}{16}\times \dfrac{1}{2}$

$\dfrac{Q_A}{Q_B}=0.031$

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Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.340 m and carries a current

sladkih [1.3K]

Answer:

The y-value of the line in the xy-plane where the total magnetic field is zero $U = 0.1355 \ m$

Explanation:

From the question we are told that

The distance of wire one from two along the y-axis is y = 0.340 m

The current on the first wire is $I_1 = (27.5i) A$

The force per unit length on each wire is $Z = 295 \mu N/m = 295*10^{-6} N/m$

Generally the force per unit length is mathematically represented as

$Z = \frac{F}{l} = \frac{\mu_o I_1I_2}{2\pi y}$

=> $\frac{\mu_o I_1I_2}{2\pi y} = 295$

Where $\mu_o$ is the permeability of free space with a constant value of $\mu_o = 4\pi *10^{-7} \ N/A2$

substituting values

$\frac{ 4\pi *10^{-7} 27.5 * I_2}{2\pi * 0.340} = 295 *10^{-6}$

=> $I_2 = 18.23 \ A$

Let U denote the line in the xy-plane where the total magnetic field is zero

So

So the force per unit length of wire 2 from line U is equal to the force per unit length of wire 1 from line (y - U)

So

$\frac{\mu_o I_2 }{2 \pi U} = \frac{\mu_o I_1 }{2 \pi(y - U) }$

substituting values

$\frac{ 18.23 }{ U} = \frac{ 27.5 }{(0.34 - U) }$

$6.198 -18.23U = 27.5U$

$6.198=45.73U$

$U = 0.1355 \ m$

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2 years ago

Two spherical objects have masses of 200 kg and 500 kg. Their centers are separated by a distance of 25 m. Find the gravitationa

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Answer:

1.07 x 10⁻⁸N

Explanation:

Given parameters:

Mass 1 = 200kg

Mass 2 = 500kg

Distance of separation = 25m

Unknown:

Gravitational attraction between the two bodies = ?

Solution:

To solve this problem, we use the equation of the universal gravitation;

F = $\frac{G mass 1 x mass 2}{r^{2} }$

G is the universal gravitation constant = 6.67 x 10⁻¹¹Nm²kg⁻²

r is the distance

Now insert the parameters and solve;

F = $\frac{6.67 x 10^{-11} x 200 x 500 }{25^{2} }$ = 1.07 x 10⁻⁸N

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1 year ago

An archer fires an arrow, which produces a muffled "thwok" as it hits a target. If the archer hears the "thwok" exactly 1 s afte

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Answer:

35,79 meters

Explanation:

So, we got an archer, and we got a target. Lets call the distance between this two d.

Now, the archer fires the arrow, that, in a time $t_{arrow}$ travels the distance d with a speed $v_{arrow}$ of 40 m/s and hits the target. We can see that the equation will be:

$v_{arrow} * t_{arrow} = d\\ \\40 \frac{m}{s} * t_{arrow} = d$

Immediately after this, the arrow produces a muffled sound, which will travel the distance d at 340 m/s in a time $t_{sound}$ . Obtaining :

$v_{sound} * t_{sound} = d\\ \\340 \frac{m}{s} * t_{sound} = d$ .

Finally, the sound reaches the archer, exactly 1 second after he fired the bow, so:

$t_{arrow} + t _{sound} = 1 s$ .

This equation allows us to write:

$t _{sound} = 1 s - t_{arrow}$ .

Plugging this relationship in the distance equation for the sound:

$340 \frac{m}{s} * t_{sound} = d \\ \\ 340 \frac{m}{s} * (1 s- t_{arrow}) = d$ .

Now, we can replace d from the first equation, and obtain:

$40 \frac{m}{s} * t_{arrow} = d \\ 40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * (1 s- t_{arrow})$ .

Now, we can just work a little bit:

$40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * 1 s - 340 \frac{m}{s} * t_{arrow} \\ \\ 40 \frac{m}{s} * t_{arrow} + 340 \frac{m}{s} * t_{arrow} = 340 m \\ \\ 380 \frac{m}{s} * t_{arrow} = 340 m \\ \\ t_{arrow} = \frac{340 m}{380 \frac{m}{s}} \\ \\ t_{arrow} = 0.8947 s$ .

Now, we can just plug this value into the first equation:

$40 \frac{m}{s} * t_{arrow} = d$

$40 \frac{m}{s} * 340/380 s = 35,79 s = d$

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2 years ago

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bazaltina [42]

Answer:

The correct option is (B).

Explanation:

The Kepler's third law of motion gives the relationship between the orbital time period and the distance from the semi major axis such that,

$T^2\propto a^3\\\\T^2=ka^3$

It is mentioned that, an asteroid with an orbital period of 8 years. So,

$(8)^2=ka^3\\\\64=ka^3\\\\a=(64)^{\dfrac{1}{3}}\\\\a=4\ AU$

So, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to 4 astronomical units.

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2 years ago

The study of alternating electric current requires the solutions of equations of the form i equals Upper I Subscript max Baselin

KiRa [710]

Answer:

Explanation:

i = Imax sin2πft

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Put the give values in the equation above

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sin 2πft = .9

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2 years ago