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Ugo [173]
3 years ago
8

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

Physics
1 answer:
Oliga [24]3 years ago
6 0

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

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What is one characteristic of the convection inside the sun?
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3 years ago
If a negative charge is released in a uniform electric field move
Goryan [66]

Answer:

Explanation:

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8 0
3 years ago
A man stands on the roof of a 15.0-m-tall building and throws a rock with a speed of 30.0 m>s at an angle of 33.0%1b above th
Brilliant_brown [7]
The motion described here is a projectile motion which is characterized by an arc-shaped direction of motion. There are already derived equations for this type of motions as listed:

Hmax = v₀²sin²θ/2g
t = 2v₀sinθ/g
y = xtanθ + gx²/(2v₀²cos²θ)

where

Hmax = max. height reached by the object in a projectile motion
θ=angle of inclination
v₀= initial velocity
t = time of flight
x = horizontal range
y = vertical height

Part A. 

Hmax = v₀²sin²θ/2g = (30²)(sin 33°)²/2(9.81)
Hmax = 13.61 m

Part B. In this part, we solve the velocity when it almost reaches the ground. Approximately, this is equal to y = 28.61 m and x = 31.91 m. In projectile motion, it is important to note that there are two component vectors of motion: the vertical and horizontal components. In the horizontal component, the motion is in constant speed or zero acceleration. On the other hand, the vertical component is acting under constant acceleration. So, we use the two equations of rectilinear motion:

y = v₀t + 1/2 at²
28.61 = 30(t) + 1/2 (9.81)(t²)
t = 0.839 seconds

a = (v₁-v₀)/t
9.81 = (v₁ - 30)/0.839
v₁ = 38.23 m/s

Part C. 
y = xtanθ + gx²/(2v₀²cos²θ)
Hmax + 15 = xtanθ + gx²/(2v₀²cos²θ)
13.61 + 15 = xtan33° + (9.81)x²/[2(30)²(cos33°)²]
Solving using a scientific calculator,
x = 31.91 m

3 0
3 years ago
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