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algol13
3 years ago
15

The continuity equation shows that the ratio of fluid velocities within different openings is inversely proportional to the cros

s-sectional areas of those openings. Provide one example where this might be useful for finding differences in areas after measuring velocities, or in finding differences in velocities after measuring areas. Describe your example in detail making sure to talk about the different variables involved.
Physics
1 answer:
Stolb23 [73]3 years ago
7 0

Answer:

A typical example of where the continuity equation can be applied, is in finding the cross sectional area of a laminar falling stream of water from a pipe at an elevation, at the point where it hits the ground.

The speed of the water through the discharge opening of the pipe can be measured with a flow-meter, and the cross sectional opening of the pipe is known.

As the water leaves the pipe, and falls towards the ground, the water stream is accelerated under gravity, and the speed of the water stream increase. The increase in the speed of the water stream causes the cross sectional area to decrease in obedience of the continuity law and equation. If the speed of can be measured, then the area of the water stream just before the water touches the ground  can be calculated from the continuity equation.

If the speed of the water stream can not be measured, then it can be calculated using Newton' equation of motion

v^{2} = u^{2} + 2gz

v = \sqrt{u^{2} + 2gz}

where

v = the velocity of the water stream at the point where it hits the ground

u = the velocity of the water at the pipe discharge (measured with a flow meter in the pipe)

g = acceleration due to gravity

z = the elevation of the pipe above the ground.

After calculating the speed of the water stream at the point where it hits the ground, the cross sectional area of the the stream can be calculated from the continuity equation.

uA = va

where

u and v are the velocities of the water stream at the pipe discharge and at the ground respectively.

A = the cross sectional area of the pipe

a = the area of the water stream before it hits the ground.

the equation is simplified as

a = \frac{uA}{v}

You might be interested in
Un tren emplea cierto tiempo en recorrer 240 km. Si la velocidad hubiera sido 20 km por hora mas que la que llevaba hubiera tard
podryga [215]

Answer:

A train takes some time to travel 240 km. If the speed had been 20 km per hour more than the one it was carrying, it would have taken 2 hours less to travel this distance. In what time did he cover the 240 km

Explanation:

Given that,

A train travelled a distance of 240km

Let the initial speed be

S_1 = x km/hr

Let assume the time spent on the first journey is

t_1 = a

Now if he increase the speed to

S_2 = (x + 20) km/hr

Then, he would have take 2hrs less time

Then, time t_2 = a - 2

The common data fore the two journey is the distance

Speed = distance / time

For the first stage

S_1 = d / t_1

d = S_1 × a

d = x × a

240 = x•a

x = 240 / a Equation 1

For stage two

d = S_2 × t_2

d = (x+20) × (a - 2)

240 = (x+20) × (a - 2). Equation 2

Substitute equation 1 into 2

240 = (240/a + 20) × (a -2)

240 = 240 - 480/a + 20a - 40

240 - 240 + 40 = - 480/a + 20a

240 - 240 + 40 = (-480 + 20a²) / a

40 = (-480 + 20a²) / a

40a = -480 + 20a²

20a² - 40a -480 = 0

Divided through by 20

a² - 2a - 24 = 0

a² + 4a - 6a - 24 = 0

a(a+4) -6(a+4) = 0

(a-6)(a+4) = 0

(a-6) = 0 or (a+4) = 0

So, a = 6 or a = -4

The time cannot be negative, then, the time is a = 6hours

So, t_1 = a = 6hours,

So, the time used in the first journey is 6hours

So, in the second journey the time use is 2hours less than the first journey

Then, t_2 = 6 - 2 = 4 hours

t_1 = 6 hours

t_2 = 4 hours

Spanish

Un tren recorrió una distancia de 240 km.

Deje que la velocidad inicial sea

S_1 = x km / h

Supongamos que el tiempo dedicado al primer viaje es

t_1 = a

Ahora si aumenta la velocidad a

S_2 = (x + 20) km / h

Entonces, habría tomado 2 horas menos de tiempo

Entonces, el tiempo t_2 = a - 2

Los datos comunes para los dos viajes son la distancia.

Velocidad = distancia / tiempo

Para la primera etapa

S_1 = d / t_1

d = S_1 × a

d = x × a

240 = x • a

x = 240 / a Ecuación 1

Para la etapa dos

d = S_2 × t_2

d = (x + 20) × (a - 2)

240 = (x + 20) × (a - 2). Ecuación 2

Sustituye la ecuación 1 en 2

240 = (240 / a + 20) × (a -2)

240 = 240 - 480 / a + 20a - 40

240 - 240 + 40 = - 480 / a + 20a

240 - 240 + 40 = (-480 + 20a²) / a

40 = (-480 + 20a²) / a

40a = -480 + 20a²

20a² - 40a -480 = 0

Dividido entre 20

a² - 2a - 24 = 0

a² + 4a - 6a - 24 = 0

a (a + 4) -6 (a + 4) = 0

(a-6) (a + 4) = 0

(a-6) = 0 o (a + 4) = 0

Entonces, a = 6 o a = -4

El tiempo no puede ser negativo, entonces, el tiempo es a = 6 horas

Entonces, t_1 = a = 6 horas,

Entonces, el tiempo utilizado en el primer viaje es de 6 horas

Entonces, en el segundo viaje, el uso del tiempo es 2 horas menos que el primer viaje

Entonces, t_2 = 6 - 2 = 4 horas

t_1 = 6 horas

t_2 = 4 horas

5 0
3 years ago
The PVT behavior of a certain gas is described by the equation of state: P(V − b) = RT where b is a constant. If in addition CV
alexdok [17]

Answer:

shown in the attachment

Explanation:

The detailed step by step and necessary mathematical application is as shown in the attachment.

6 0
4 years ago
The Trojan asteroids are found:_________ a) orbiting around the Kuiper Belt body Hector. b) beyond Neptune, with orbits similar
xenn [34]

Answer:

The correct option is;

c) sixty degrees ahead or behind Jupiter, sharing its orbit about the Sun.

Explanation:

The Trojan asteroids are the Jupiter trojans consists of asteroid that are on the same orbit as Jupiter while moving around the Sun. The Trojans can be located at the points Lagrange points L4 and L5, which are 60° ahead and 60°  behind Jupiter's orbit respectively.

The first Trojan asteroid to be detected was 588 Achilles by Max Wolf in 1906. At at October, the total number of the identified Trojan asteroid was  7,040.

5 0
3 years ago
Read 2 more answers
The membrane that surrounds a certain type of living cell has a surface area of 5.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A
kotykmax [81]

Answer:

2.1\times 10^{-12} c

Explanation:

We are given that

Surface area of membrane=5.3\times 10^{-9} m^2

Thickness of membrane=1.1\times 10^{-8} m

Assume that membrane behave like a parallel plate capacitor.

Dielectric constant=5.9

Potential difference between surfaces=85.9 mV

We have to find the charge resides on the outer surface of membrane.

Capacitance between parallel plate capacitor is given by

C=\frac{k\epsilon_0 A}{d}

Substitute the values then we get

Capacitance between parallel plate capacitor=\frac{5.9\times 8.85\times 10^{-12}\times 5.3\times 10^{-9}}{1.1\times 10^{-8}}

C=0.25\times 10^{-12}F

V=85.9 mV=85.9\times 10^{-3}

Q=CV

Q=0.25\times 10^{-12}\times 85.9\times 10^{3}=2.1\times 10^{-12} c

Hence, the charge resides on the outer surface=2.1\times 10^{-12} c

5 0
3 years ago
Read 2 more answers
A spring is compressed 0.035 m inside a dart gun. (K=500 N/m). The spring has elastic energy. Calculate it.
NARA [144]

The elastic potential energy of the spring is 0.31 J

Explanation:

The elastic potential energy of a spring is given by

E=\frac{1}{2}kx^2

where

k is the spring constant

x is the compression/stretching of the spring

For the spring in this problem, we have:

k = 500 N/m (spring constant)

x = 0.035 m (compression)

Substituting, we find the elastic potential energy:

E=\frac{1}{2}(500)(0.035)^2=0.31 J

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

6 0
3 years ago
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