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konstantin123 [22]

3 years ago

15

Consider the flow of an incompressible Newtonian fluid between two parallel plates that are 4 mm apart. If the upper plate moves

to the right with u1 = 5 m/s, while the bottom one moves to the left with u2 = 1.5 m/s, what would be the net flow rate at a cross section between the two plates? Take the plate width to be 5.5 cm and the distance between the plates to be 4 mm.

1 answer:

LekaFEV [45]3 years ago

3 0

Answer:

$3.85*10^{-5}m^3/a$

Explanation:

We need to find the average velocity of each plate, $u_1$ and $u_2$ .

So substituting,

$\bar{V_1} = \frac{1}{2}*u_1=\frac{1}{2}*5m/s=2.5m/s$

$\bar{V_2} = \frac{1}{2}*u_2 = \frac{1}{2}*1.5m/s=0.75m/s$

The net velocity is: $\bar{V_1}-\bar{V_2}=2.5-0.75=1.75m/s$

The flow rate = $\bar{V}_{net}A = 1.75m/s * 5.5*10^{-3} * 4*10^{-3}$

The flow rate $= \bar{V}_{net}A = 3.85*10^{-5}m^3/a$

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Answer:

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3 years ago

steel wire 8m long and 4mm in diameter is fixed to two rigid b. [1] supports. Calculate the increase in tension when the tempera

marin [14]

Answer:

301.6 N

Explanation:

The length of the wire L₀ = 8 m and its diameter, d = 4 mm = 4 × 10⁻³ m. Since its temperature drops by 10°C, it will have a change in length ΔL = L₀αΔθ where α = linear expansivity of steel, a 12 × 10⁻⁶ /K, and Δθ = temperature change = -10°C = -10 K(negative since it is a drop)

So, the strain, ε = ΔL/L₀ = αΔθ = 12 × 10⁻⁶ /K × 10 K = 12 × 10⁻⁵

Now the Young's modulus of steel, Y = σ/ε where σ = stress = T/A where T = increase in tension in steel wire and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 4 × 10⁻³ m and ε = strain = 12 × 10⁻⁵

So, σ = Yε

Since Y = 2 × 10¹¹ N/m².

Substituting the values of the variables into the equation, we have

σ = Yε

σ = 2 × 10¹¹ N/m² × 12 × 10⁻⁵

σ = 24 × 10⁶ N/m²

Since σ = T/A

T = σA

T = σπd²/4

Substituting the values of the variables into the equation, we have

T = σπd²/4

T = 24 × 10⁶ N/m² × π × (4 × 10⁻³ m)²/4

T = 24 × 10⁶ N/m² × π × 16 × 10⁻⁶ m²/4

T = 24 × 10⁶ N/m² × π × 4 × 10⁻⁶ m²

T = 96 N × π

T = 301.59 N

T ≅ 301.6 N

So, the increase in tension in the steel wire is 301.6 N

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3 years ago

An AED should only be used if a person is too tired to continue with chest compressions.

malfutka [58]

The given statement "An AED should only be used if a person is too tired to continue with chest compressions" is false.

Answer: Option B

<u>Explanation:</u>

Most people think that they are very tired after applying the compress for 2-3 minutes. When the compressor is tired, it tends to compress more slowly. For this reason, rescuers are advised to do compressions in every 2 min to prevent fatigue and optimise the compressions quality. This should only be done if the person does not show signs of life or is dead, does not respond and breathes normally.

For example, consider you come across the victim in a narrow place and you have two helpers: Rescuer 2 arrives with the AED (automated external defibrillator) and puts it on the opposite side from Rescuer 1, who does chest compressions. Rescuer 2 turns on the AED and fixes the electrodes to the victim's chest, connecting wires to the AED if necessary. Rescuer 1 can continue CPR (Cardiopulmonary resuscitation) by placing electrodes until the victim's heart rate has been analysed.

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3 years ago

A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co

algol [13]

Answer:

$\displaystyle X_T=66.6\ km$

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

$\displaystyle V_f=V_o+a\ t$

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of $0.987 m/s^2$ for 182 seconds. Then it brakes with a constant acceleration of $-0.321 m/s^2$ until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

$\displaystyle X=V_o\ t+\frac{a\ t^2}{2}$

Our data is

$\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec$

Let's compute the first distance X1

$\displaystyle X_1=0+\frac{0.987\times 182^2}{2}$

$\displaystyle X_1=16,346.7\ m$

Now, we find the speed at the end of the first period of time

$\displaystyle V_{f1}=0+0.987\times 182$

$\displaystyle V_{f1}=179.6\ m/s$

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

$\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0$

$\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}$

$\displaystyle t=559.5\ sec$

Computing the second distance

$\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}$

$\displaystyle X_2=50,243.2\ m$

The total distance is

$\displaystyle X_t=x_1+x_2=16,346.7+50,243.2$

$\displaystyle X_t=66,589.9\ m$

$\displaystyle \boxed{X_T=66.6\ km}$

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4 years ago

Which Best describes a characteristic of the jet stream?

Ipatiy [6.2K]

Answer:

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Explanation:

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3 years ago