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vodka [1.7K]
3 years ago
14

Which type of valve is commonly used on private water supply systems and the words open or shut appear in a window as the valve

approaches one position or the other?
Physics
1 answer:
podryga [215]3 years ago
4 0

Answer:

Post indicator valves

Explanation:

Post indicator valves were first made and patented by its inventors Crookham and Lee in 1972 for Clow Corp., a pipe manufacturing company. In 1974, they were publicly installed for the first time.

<em>PIV, indicator post, I post, IP, post indicator, or wall post, a post indicator valve</em> is a valve assembly used for the purpose of underground piping for water supply to fire protection systems such as sprinklers, sprays, foams, deluges, etc. They have a lockable actuator which looks like a red metal steering wheel or a wrench/handle on top of a post. It consists of a glass or plastic window, which indicates whether the valve is open or shut.

<u>PIVs can be of 3 types: </u>

<u />

Ground Post: This is an indicator post that has its base buried in the ground and typically has a lockable handle or clamp. It is used when the valve is installed out of the building.

Wall Post: This is a short indicator post, usually horizontally bound through a flange to the wall of the building. This post can be operated by a handwheel.

Wall Post: This is a short indicator post, usually bolted horizontally to the wall of the building through a flange. This post is operated through a handwheel. This is used when the water supply passes through a cavity of the wall.

Pedestal Post: This post looks like a ground sign. Furthermore, instead of being embedded in the earth, it is bolted through a flange to a horizontal surface, usually a concrete base. It also operates through a handle or wrench that can be locked. It is helpful when the valve is situated inside a building and the water supply is stored

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The correct option is C.
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3 years ago
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If the distance to a point source of sound is doubled, by what multiplicative factor does the intensity change?
alina1380 [7]

If the distance to a point source of sound is doubled, by a multiplicative factor of 4, the intensity changes.

Intensity of sound is the sound which is perpendicular to sound wave propogation per unit area. It is dependent on the Surface of source sound.

Intensity is the Power per unit area. Its SI unit is Watt/m².

As we move away from a source of sound, the sound starts to diminish. This is due to the decreasing sound intensity with distance.

It can also be understood by the fact that on increasing distance, the Power radiated by the source spreads over a larger area. Hence, the Intensity decreases gradually.

Since, Intensity is proportional to the square of the distance.

Hence, on doubling the distance, Intensity reduces to one fourth of the initial intensity or reduces by a multiplicative factor of 4.

Learn more about Intensity here, brainly.com/question/17583145

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A certain ultrasound device can measure a fetal heart rate as low as 50 beats per minute. This corresponds to the surface of the
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The. Machine must detect a shift of

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Explanation:

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={ ( Vsound +V/ V sound -V) -1}f emitted

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3 years ago
Lagrangian mechanics. Determine the equations of motion for a particle of mass m constrained to move on the surface of a cone in
maria [59]

Answer:

Explanation:

Hi!

In order to obtain the Lagrangian of the system we must first write the Kinetic and Potential Energies. Lets orient our axes such that the axis of the cone coincide with the z axis. In cilindrical coordinates we have

v^{2} = \frac{dr}{dt}^{2}  +r^{2} \frac{d\theta }{dt} ^{2} +\frac{dz}{dt} ^{2} - (1)

But, since the particle is constrained to move on the surface of the cilinder, we have the following relation between r and z:

\frac{r}{z}=tan(45)

or:

z = r cot(45) - (2)

and:

\frac{dz}{dt} = \frac{dr}{dt} cot(45)

replacing (2) in (1) we obtain:

v^{2} = \frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2}  - (3)

Now the kinetic energy is given as:

T = \frac{1}{2}m(\frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2}) - (4)

And the potential energy is given by:

V = -mgz = -mgr cot(45)

So the Langrangian is given by:

L = T - V= \frac{1}{2}m(\frac{dr}{dt}^{2}(1+cot(45)+r^{2})\frac{d\theta }{dt} ^{2}) + mgr cot(45)

And the equations of motion are:

For θ

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For r

\frac{d}{dt}(m\frac{dr}{dt}(1+cot(45) )= mgcot(45)+mr\frac{d\theta}{dt} ^{2}\\m\frac{d^{2} r}{dt^{2} }(1+cot(45)= mgcot(45)+mr\frac{d\theta}{dt} ^{2}

Obtained from the Euler-Langrange equations

Here the conserved quantity is given by the first equation of motion, namely:

mr\frac{d\theta}{dt}=c

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