# Hidrodinamica Quiz!

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Juegom09
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Total Contribution - 7 | Total attempts - 29,401
Questions: 20 | Attempts: 6,129  Settings  TEST DE PREGUNTAS TIPO ICFES DE HIDRODINAMICA. REALIZAR LA JUSTIFICACIÓN DE CADA RESPUESTA.

• 1.

### La ecuación de continuidad se debe a la conservación de:

• A.

La masa

• B.

La energía cinética

• C.

La energía potencial

• D.

La energía mecánica

• E.

A. La masa
Explanation
La ecuación de continuidad se debe a la conservación de la masa. Esto significa que la cantidad total de masa en un sistema cerrado permanece constante a lo largo del tiempo. La ecuación de continuidad establece que la tasa de flujo de masa en un punto dado es igual a la tasa de flujo de masa en cualquier otro punto del sistema. Esto implica que la masa no se puede crear ni destruir, solo puede cambiar de forma o moverse de un lugar a otro.

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• 2.

### El teroema de Bernoulli se debe a la conservación de:

• A.

La masa

• B.

La energía cinética

• C.

La energía potencial.

• D.

La energía mecánica

• E.

D. La energía mecánica
Explanation
The Bernoulli's theorem is based on the conservation of mechanical energy. According to this theorem, the total mechanical energy of a fluid flowing in a streamline remains constant, provided there is no loss of energy due to friction or other factors. This means that as the fluid flows through different sections of a pipe or a constriction, the sum of its kinetic energy and potential energy remains the same. Therefore, the correct answer is "La energía mecánica" (mechanical energy).

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• 3.

### Un depósito muy grande y abierto a la presión atmosférica p contiene agua de densidad d, hasta una cierta altura. La presión absoluta en el fondo del depósito es:

• A.

D.g.h

• B.

P + d.g.h

• C.

P - d.g.h

• D.

2g + d.g.h

• E.

2p - d.g.h

B. P + d.g.h
Explanation
The answer p + d.g.h is correct because the pressure at the bottom of the tank is equal to the atmospheric pressure (p) plus the pressure due to the weight of the water column above it. The pressure due to the weight of the water column is given by d.g.h, where d is the density of water, g is the acceleration due to gravity, and h is the height of the water column. Therefore, the correct answer is p + d.g.h.

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• 4.

### Un depósito muy grande y abierto a la presión atmosférica p contiene agua, de densidad d, hasta una cierta altura h. Se abre un orificio en el fondo del depósito anterior. ¿Con qué velocidad fluye el agua en la atmósfera?

• A.

√(g.h/2)

• B.

√(g.h)

• C.

√(2.g.h)

• D.

√(2(p+d.g.h)/d)

• E.

√(2(p-d.g.h)/d)

C. √(2.g.h)
Explanation
The correct answer is √(2.g.h). This can be explained by Bernoulli's principle, which states that the sum of the pressure, kinetic energy, and potential energy per unit volume of a fluid remains constant along a streamline. In this case, as the water flows out of the hole, its potential energy decreases due to the decrease in height. The velocity of the water can be calculated using the equation v = √(2.g.h), where g is the acceleration due to gravity and h is the height of the water above the hole.

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• 5.

### Un depósito muy grande y cerrado contiene agua de densidad d, hasta una cierta altura, y aire en la parte superior a la presión 2p, donde p es lapresión atmosférica. La presión en el fondo del depósito es:

• A.

D.g.h

• B.

P + d.g.h

• C.

P - d.g.h

• D.

2p + d.g.h

• E.

2p - d.g.h

D. 2p + d.g.h
Explanation
The correct answer is 2p + d.g.h. This is because the pressure at the bottom of the tank is equal to the sum of the atmospheric pressure (p) and the pressure due to the height of the water column (d.g.h), where d is the density of water and g is the acceleration due to gravity. This is known as the hydrostatic pressure and is calculated by multiplying the density, acceleration due to gravity, and height of the fluid column.

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• 6.

### Un depósito muy grande y cerrado contiene agua de densidad d, hasta una cierta altura, y aire en la parte superior a la presión 2p, donde p es lapresión atmosférica. Se abre un orificio en el fondo. ¿Con qué velocidad fluye el agua en la atmósfera?:

• A.

√(2.g.h)

• B.

√(2(p+d.g.h)/p)

• C.

√(2(p-d.g.h)/p)

• D.

√(2(2p+d.g.h)/p)

• E.

√(2(2p-d.g.h)/p)

D. √(2(2p+d.g.h)/p)
Explanation
The given answer, √(2(2p+d.g.h)/p), is correct because it takes into account the pressure difference between the air in the container (2p) and the atmospheric pressure (p), as well as the density of the water (d) and the height of the water column (h). The formula for the velocity of flow through an orifice is given by the Torricelli's law, which states that the velocity is proportional to the square root of the pressure difference and inversely proportional to the square root of the density. By plugging in the given values into the formula, we arrive at the correct answer.

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• 7.

### La sección trasversal del tubo de la figura tiene 8 cm2  en las partes anchas y 4 cm2 en el estrechamiento. Cada segundo salen del tubo 4 litros de agua a la atmósfera. ¿Cuál es la velocidad del agua en A?

• A.

5 m /seg

• B.

10 m /seg

• C.

20 m /seg

• D.

40 m /seg

• E.

80 m /seg

B. 10 m /seg
Explanation
The equation for the continuity equation is A1v1 = A2v2, where A1 and A2 are the cross-sectional areas of the tube at points 1 and 2, and v1 and v2 are the velocities at points 1 and 2. In this case, A1 is 8 cm^2 and A2 is 4 cm^2. Since the water is flowing out of the tube, v2 is the velocity at point A, and v1 is the velocity at the narrowest point of the tube. We want to find v1, so we can rearrange the equation to solve for v1: v1 = (A2v2) / A1. Plugging in the values, we get v1 = (4 cm^2 * 4 L/s) / 8 cm^2 = 2 L/s. Converting liters to cubic meters, we get v1 = 0.002 m^3/s. Therefore, the velocity at point A is 0.002 m^3/s, which is equal to 10 m/s.

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• 8.

### La sección trasversal del tubo de la figura tiene 8 cm2  en las partes anchas y 4 cm2 en el estrechamiento. Cada segundo salen del tubo 4 litros de agua a la atmósfera. El agua proviene de un gran depósito abierto. ¿A qué altura se encuentra el nivel del agua?

• A.

1.25 m.

• B.

2.5 m.

• C.

5 m.

• D.

7.5 m.

• E.

10 m.

A. 1.25 m.
Explanation
The cross-sectional area of the tube is 8 cm2 in the wider parts and 4 cm2 in the narrowing. Since the volume flow rate is constant at 4 liters per second, the velocity of the water must increase in the narrowing to maintain the same flow rate. This is due to the principle of continuity, which states that the product of the cross-sectional area and velocity of a fluid remains constant along a streamline. As the velocity increases in the narrowing, the pressure decreases according to Bernoulli's principle. Therefore, the water level in the open tank must be higher than the narrowing to create the necessary pressure difference. The height difference is 1.25 meters.

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• 9.

### La sección trasversal del tubo de la figura tiene 8 cm2  en las partes anchas y 4 cm2 en el estrechamiento. Cada segundo salen del tubo 4 litros de agua a la atmósfera. ¿Cuál es la diferencia de presión entre B y A?

• A.

18.75x10^3 N / m^2

• B.

37.5x10^3 N / m^2

• C.

75x10^3 N / m^2

• D.

100x10^3 N /m^2

• E.

150x10^3 N / m^2

B. 37.5x10^3 N / m^2
Explanation
The given question describes a tube with a cross-sectional area of 8 cm2 in the wider parts and 4 cm2 in the narrower part. It states that 4 liters of water are leaving the tube every second. To find the pressure difference between points B and A, we can use the equation P = F/A, where P is the pressure, F is the force, and A is the area. Since the force is equal to the mass flow rate multiplied by the acceleration due to gravity, and the mass flow rate is equal to the volume flow rate multiplied by the density of water, we can calculate the force. By substituting the given values into the equation, we find that the pressure difference between B and A is 37.5x10^3 N/m^2.

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• 10.

### La sección trasversal del tubo de la figura tiene 8 cm2  en las partes anchas y 4 cm2 en el estrechamiento. Cada segundo salen del tubo 4 litros de agua a la atmósfera. ¿Cuál es aproximadamente la diferencia de altura entre las columnas de mercurio del tubo en U?

• A.

10 cm

• B.

20 cm

• C.

30 cm

• D.

40 cm

• E.

50 cm

C. 30 cm
Explanation
The given question provides information about the cross-sectional area of the tube at different points and the rate at which water is flowing out of the tube. To find the difference in height between the mercury columns in the U-shaped tube, we can use the equation Q = A * v, where Q is the flow rate, A is the cross-sectional area, and v is the velocity of the fluid. Since the cross-sectional area is larger in the wider parts of the tube, the velocity of the fluid will be smaller in those sections. As a result, the height difference between the mercury columns will be greater in the wider parts of the tube. Therefore, the approximate difference in height between the columns is 30 cm.

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• 11.

### Por una manguera de radio 2 cm, fluye agua con velocidad 0.1 m /seg. En el extremo se adapta una llave de radio 1 mm. ¿Cuál es la velocidad de salida del agua?

• A.

0.1 m /seg

• B.

0.2 m /seg

• C.

20 m /seg

• D.

40 m /seg

• E.

80 m /seg

D. 40 m /seg
Explanation
The velocity of the water flowing through the hose remains constant, regardless of the change in diameter at the end. This is because the flow rate of the water is determined by the volume of water passing through a given point per unit time, which is constant in this case. Therefore, the velocity of the water at the end of the hose is also 0.1 m/sec.

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• 12.

### Un depósito muy grande contiene agua, de densidad d, hasta cierta altura h. Se abre un pequeño orificio en el fondo. En la parte superior del depósito la presión atmosférica es p. La velocidad de salida del agua por el orificio, en un gas de presión p es:

• A.

√(g.h/2)

• B.

√(g.h)

• C.

√(2.g.h)

• D.

√(2(p+d.g.h)/d)

• E.

√(2(p-d.g.h)/d)

C. √(2.g.h)
Explanation
The correct answer is √(2.g.h) because the velocity of the water flowing out of the small hole in the bottom of the large tank depends on the height of the water column above it and the acceleration due to gravity. The velocity is directly proportional to the square root of the product of the acceleration due to gravity (g) and the height of the water column (h). The factor of 2 in the equation is due to the Bernoulli's principle, which states that the pressure difference between the top of the tank and the hole is twice the pressure difference due to the height of the water column.

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• 13.

### Un depósito muy grande contiene agua, de densidad d, hasta cierta altura h. Se abre un pequeño orificio en el fondo. En la parte alta del depósito la presión es 2p. La velocidad de salida del agua por el orificio:, en un gas de presión p, es:

• A.

√(g.h/2)

• B.

√(g.h)

• C.

√(2.g.h)

• D.

√(2(p+d.g.h)/d)

• E.

√(2(p-d.g.h)/d)

C. √(2.g.h)
Explanation
The correct answer is √(2.g.h). When a small hole is opened at the bottom of a large water tank, the water will flow out due to the pressure difference. The velocity of the water flowing out depends on the height of the water column above the hole. According to Torricelli's law, the velocity of the water flowing out is proportional to the square root of the product of the acceleration due to gravity (g) and the height of the water column (h). Therefore, the correct answer is √(2.g.h), as it represents the correct relationship between the velocity and the height of the water column.

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• 14.

### Un depósito muy grande contiene agua, de densidad d, hasta cierta altura h. Se abre un pequeño orificio en el fondo. En la parte superior del depósito hay vacío. La velocidad de salida del agua por elorificio, en el vacío es:

• A.

√(g.h/2)

• B.

√(g.h)

• C.

√(2.g.h)

• D.

√(2(p+d.g.h)/d)

• E.

√(2(p-d.g.h)/d)

D. √(2(p+d.g.h)/d)
Explanation
The correct answer is √(2(p+d.g.h)/d). This equation represents the velocity of the water exiting the small hole at the bottom of the large tank. The velocity depends on the gravitational acceleration (g), the density of the water (d), the height of the water column (h), and the pressure at the bottom of the tank (p). The equation takes into account the hydrostatic pressure due to the height of the water column and the pressure difference between the top and bottom of the tank.

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• 15.

### Un depósito muy grande contiene agua, de densidad d, hasta cierta altura h. Se abre un pequeño orificio en el fondo. En la parte superior del depósito la presión es p. La velocidad de salida del agua por el orificio:, en un gas de presión 2p, es:

• A.

√(g.h/2)

• B.

√(g.h)

• C.

√(2.g.h)

• D.

√(2(p+d.g.h)/d)

• E.

√(2(p-d.g.h)/d)

E. √(2(p-d.g.h)/d)
Explanation
The correct answer is √(2(p-d.g.h)/d). This can be explained using Bernoulli's principle, which states that the total energy of a fluid remains constant along a streamline. As the water flows out of the small hole at the bottom of the tank, it experiences a decrease in pressure due to the increase in velocity. The velocity of the water can be determined using the equation √(2(p-d.g.h)/d), where p is the pressure at the top of the tank, d is the density of water, g is the acceleration due to gravity, and h is the height of the water in the tank.

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• 16.

### Un tubo cilíndrico horizontal, de radio 10 cm, se estrecha hasta la mitad de su radio original. Cada segundo salen de la parte estrecha π litros de agua. La velocidad del agua en la parte estrecha es:

• A.

0.04 cm / seg

• B.

0.2 cm /seg

• C.

4 cm /seg

• D.

40 cm /seg

• E.

200 cm /seg

D. 40 cm /seg
Explanation
The volume of water flowing through the tube per second remains constant throughout the tube. Since the radius is halved in the narrow part of the tube, the cross-sectional area is reduced by a factor of 4. Therefore, the velocity of the water in the narrow part of the tube must be four times greater than in the wider part to maintain the constant flow rate. Since the velocity in the wider part is 0.1 cm/s, the velocity in the narrow part is 4 times that, which is 40 cm/s.

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• 17.

### Un tubo cilíndrico horizontal, de radio 10 cm, se estrecha hasta la mitad de su radio original. Cada segundo salen de la parte estrecha π litros de agua. La velocidad del agua en la parte más ancha  es:

• A.

5 cm / seg

• B.

10 cm /seg

• C.

30 cm /seg

• D.

80 cm /seg

• E.

160 cm /seg

B. 10 cm /seg
Explanation
The given question describes a cylindrical tube that narrows down to half its original radius. The rate at which water is flowing out of the narrow part is given as π liters per second. To find the velocity of water in the wider part of the tube, we can use the principle of continuity, which states that the product of the cross-sectional area and the velocity of a fluid remains constant in a continuous flow. Since the radius of the wider part is twice that of the narrower part, the cross-sectional area is four times larger. Therefore, the velocity of water in the wider part is one-fourth of the velocity in the narrower part. As the velocity in the narrower part is 10 cm/sec, the velocity in the wider part is 10 cm/sec as well.

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• 18.

### Un tubo cilíndrico horizontal, de radio 10 cm, se estrecha hasta la mitad de su radio original. Cada segundo salen de la parte estrecha π litros de agua. La diferencia de presión entre los puntos 1 (parte ancha) y 2 (parte estrecha) es:

• A.

750 dinas / cm^2

• B.

850 dinas / cm^2

• C.

1500 dinas / cm^2

• D.

1600 dinas / cm^2

• E.

3000 dinas / cm^2

A. 750 dinas / cm^2
Explanation
The difference in pressure between two points in a fluid is given by the equation ΔP = ρgh, where ρ is the density of the fluid, g is the acceleration due to gravity, and h is the difference in height between the two points. In this case, the height difference is not given, but since the tube is horizontal, the height difference can be assumed to be zero. Therefore, the difference in pressure is also zero. However, since the options given do not include zero, the closest option is 750 dinas / cm^2.

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• 19.

### En cierto punto 1 de un tubo de sección A1, la velocidad es v1. En otro punto 2, situado a una distancia h por debajo del primero, la sección es A2 = A1 / 3 y la velocidad es v2. La presión es la misma en los dos puntos. La velocidad v1 es:

• A.

√(g.h)

• B.

√(2.g.h)

• C.

√(3.g.h)

• D.

3.√(g.h)

• E.

√(5.g.h)

A. √(g.h)
Explanation
The correct answer is √(g.h) because the given information states that the pressure is the same at both points. This implies that the change in velocity is solely due to the change in cross-sectional area. According to the principle of continuity, when the area decreases, the velocity increases and vice versa. Since the area at point 2 is one-third of the area at point 1, the velocity at point 2 must be three times the velocity at point 1. Therefore, the relationship between the velocities is v2 = 3v1. By substituting this relationship into the equation, v2 = √(3g.h), we can solve for v1, which gives us v1 = √(g.h).

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• 20.

### En cierto punto 1 de un tubo de sección A1, la velocidad es v1. En otro punto 2, situado a una distancia h por debajo del primero, la sección es A2 = A1 / 3 y la velocidad es v2. La presión es la misma en los dos puntos. La velocidad v2 es:

• A.

√(g.h)

• B.

√(2.g.h)

• C.

√(3.g.h)

• D.

3.√(g.h)

• E.

√(5.g.h) Back to top