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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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Hello!
Montano1993 [528]

It is true that the  light is 15.000 more dangerous than the radiation of a microwave.

<h3>What is the wavelength?</h3>

The wavelength shows the extent or how far the wave travels. Now we know that the energy of the wave can be use to find out how much dangerous the wave is.

Now;

1.6 * 10^-19 J = 1eV

x J = 1.8 eV

x = 1.8 eV * 1.6 * 10^-19 J /1eV

x = 2.88 * 10^-19 J

Now if the energy of the microwaves is 1.2 x 10^-4 J, then it follows that;

2.88 * 10^-19 J/ 1.2 x 10^-4 J,

= 2.4 * 10^15

Hence, it is true that the  light is 15.000 more dangerous than the radiation of a microwave.

Learn more about microwave:brainly.com/question/15708046

#SPJ1

5 0
2 years ago
Eric threw a baseball 20 meters in 0.5 seconds. What was the average speed of the baseball to the nearest hundredths of a m/sec.
anzhelika [568]

Answer:

40m/sec

Explanation:

speed = distance ÷ time

speed = 20 ÷ 0.5

speed = 40 m/sec

4 0
3 years ago
Read 2 more answers
An upright spring with a 96g mass on it is compressed 2 cm. When
Alexeev081 [22]

Answer:

I only know answer A and it's 2825.28 N/m, with rounding it's 2825.5

Explanation:

Use the m*g*h=1/2*k*x^2 equation

96*9.81*60=1/2*k*2^2

5650.56=2k

5650.56/2=2825.28N/m

8 0
3 years ago
a ball is projected upward at time t = 0.00 s from a point on a roof 70 m above the ground. The ball rises, then falls and strik
grin007 [14]

Answer: 17.68 s

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:  

y=y_{o}+V_{o}t-\frac{1}{2}gt^{2} (1)  

Where:  

y=0 is the height of the ball when it hits the ground  

y_{o}=70 m is the initial height of the ball

V_{o}=82m/s is the initial velocity of the ball  

t is the time when the ball strikes the ground

g=9.8m/s^{2} is the acceleration due to gravity  

Having this clear, let's find t from (1):  

0=70m+(82m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2} (2)  

Rewritting (2):

-\frac{1}{2}(9.8m/s^{2})t^{2}+(82m/s)t+70m=0 (3)  

This is a quadratic equation (also called equation of the second degree) of the form at^{2}+bt+c=0, which can be solved with the following formula:

t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}  (4)

Where:

a=-\frac{1}{2}(9.8m/s^{2}

b=82m/s

c=70m

Substituting the known values:

t=\frac{-82 \pm \sqrt{82^{2}-4(-\frac{1}{2}(9.8)(70)}}{2a}  (5)

Solving (5) we find the positive result is:

t=17.68 s

7 0
3 years ago
How is work calculated when the force applied is not parallel to the displacement
Leno4ka [110]
Pretty sure it’s Force*Distance*Cos(theta)
7 0
3 years ago
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