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Vanyuwa [196]
3 years ago
5

How an application of atmospheric device work?example siphon​

Physics
1 answer:
Vika [28.1K]3 years ago
6 0

Answer:

A practical siphon, operating at typical atmospheric pressures and tube heights, works because gravity pulling down on the taller column of liquid leaves reduced pressure at the top of the siphon (formally, hydrostatic pressure when the liquid is not moving).

I hope it's helpful!

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Por una tubería de 0.06 m de diámetro circula agua con una velocidad desconocida, al llegar a la parte estrecha de la tubería de
Vesnalui [34]

Answer:

La velocidad con la que se desplaza el agua antes de llegar a la parte estrecha de la tubería es 1.156 \frac{m}{s}

Explanation:

La ecuación de continuidad es simplemente una expresión matemática del principio de conservación de la masa.  Este principio establece que la masa de un objeto o colección de objetos nunca cambia con el tiempo.

La ecuación de continuidad es la relación que existe entre el área y la velocidad que tiene un fluido en un lugar determinado y dice que el caudal de un fluido es constante a lo largo de un circuito hidráulico.

En otras palabras, la ecuación de continuidad se basa en que el caudal (Q) del fluido ha de permanecer constante a lo largo de toda la conducción. Cuando un fluido fluye por un conducto de diámetro variable, su velocidad cambia debido a que la sección transversal varía de una sección del conducto a otra.

Entonces, siendo el caudal es el producto de la superficie de una sección del conducto por la velocidad con que fluye el fluido,  en dos puntos de una misma tubería se cumple:

Q1=Q2

A1*v1= A2*v2

donde:

  • A es la superficie de las secciones transversales de los puntos 1 y 2 del conducto.
  • v es la velocidad del flujo en los puntos 1 y 2 de la tubería.

Siendo A=pi*r^{2} =pi*(\frac{D}{2} )^{2} =\frac{pi*D^{2} }{4} , donde pi es el número π, r es el radio del conducto y D el diámetro del conducto, entonces:

\frac{pi*D1^{2} }{4}*v1=\frac{pi*D2^{2} }{4}*v2

En este caso:

  • D1: 0.06 m
  • v1: ?
  • D2: 0.04 m
  • v2: 2.6 m/s

Reemplazando:

\frac{pi*(0.06m)^{2} }{4}*v1=\frac{pi*(0.04m)^{2} }{4}*2.6\frac{m}{s}

Resolviendo:

v1=\frac{\frac{pi*(0.04m)^{2} }{4}*2.6\frac{m}{s}}{\frac{pi*(0.06m)^{2} }{4}}

v1=\frac{(0.04m)^{2} }{(0.06m)^{2}  }*2.6\frac{m}{s}

v1= 1.156 \frac{m}{s}

<u><em>La velocidad con la que se desplaza el agua antes de llegar a la parte estrecha de la tubería es 1.156 </em></u>\frac{m}{s}<u><em></em></u>

8 0
3 years ago
This is an example of:
erica [24]

Answer:

C. both

Explanation:

kinetic energy bcs it's moving and gravitational potential energy for it's descend.

4 0
3 years ago
What type of plate tectonic activity do you think is being shown in the picture below?
blondinia [14]

Answer:

Convergent boundaries,over time,a rift valley will form.

Explanation:

The picture shows 2 plates moving apart,and a shallow "valley"is also noticeable.

Mark me brainliest if I helped:D

3 0
3 years ago
A bar-magnet with magnetic moment 2.5 Am^2 is placed in a homogeneous magnetic field (of 0.1 T that is directed along the z-axis
Angelina_Jolie [31]

Answer:

1)

Force on bar magnet  = 0

Torque on bar magnet = 0

2)

Force on bar magnet  = 0

Torque on bar magnet = 0.177 Nm

3)

Force on bar magnet  = 0

Torque on bar magnet = 0.25 Nm

Explanation:

Part 1)

net force on bar magnet in uniform magnetic field is always zero

Torque on bar magnet is given as

\tau = MBsin\theta

when bar magnet is inclined along z axis along magnetic field

then we will have

\tau = MBsin0 = 0

Part 2)

net force on bar magnet in uniform magnetic field is always zero

Torque on bar magnet is given as

\tau = MBsin\theta

when bar magnet is pointing 45 degree with z axis then we will have

\tau = MBsin45

\tau = (2.5)(0.1)sin45

\tau = 0.177 Nm

Part 3)

net force on bar magnet in uniform magnetic field is always zero

Torque on bar magnet is given as

\tau = MBsin\theta

when bar magnet is pointing 90 degree with z axis then we will have

\tau = MBsin90

\tau = (2.5)(0.1)sin90

\tau = 0.25 Nm

8 0
3 years ago
A tennis ball is dropped from 1.3 m above
Dominik [7]

Answer:

2.65m/s

Explanation:

Using the equation of motion:

v² = u²+2a∆S where

v is the final velocity

u is the initial velocity

∆S is the change in distance

a is the acceleration

Given

u = 0m/s

a = 9.8m/s²

∆S = 1.3-0.943

∆S = 0.357m

Substituting the given parameters into the formula

v² = 0²+2(9.8)(0.357)

v² = 0+6.9972

v² = 6.9972

v=√6.9972

v = 2.65m/s

Hence the velocity at which it hit the ground is 2.65m/s

4 0
3 years ago
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