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3 years ago

HIIIIIIIIII htrdcvbnjuytresdxcvbnjhytrdefghjkmnbvcdertyhjmn i'm soo bored

Engineering

2 answers:

Aloiza [94]3 years ago

8 0

Answer:

My name is Yoshikage Kira

Explanation:

My name is Yoshikage Kira. I'm 33 years old. My house is in the northeast section of Morioh, where all the villas are, and I am not married. I work as an employee for the Kame Yu department stores, and I get home every day by 8 PM at the latest. I don't smoke, but I occasionally drink. I'm in bed by 11 PM, and make sure I get eight hours of sleep, no matter what. After having a glass of warm milk and doing about twenty minutes of stretches before going to bed, I usually have no problems sleeping until morning. Just like a baby, I wake up without any fatigue or stress in the morning. I was told there were no issues at my last check-up. I'm trying to explain that I'm a person who wishes to live a very quiet life. I take care not to trouble myself with any enemies, like winning and losing, that would cause me to lose sleep at night. That is how I deal with society, and I know that is what brings me happiness. Although, if I were to fight I wouldn't lose to anyone.

sweet [91]3 years ago

3 0

I’m bored too hejdhejfnfndndnd

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The application of technology results in human-made things called

Sergio039 [100]

Answer:

Internet of things

Explanation:

This is a good example where the application of technology results are applied to human made things.

Internet of things (IOT), involves the application of one technology results–the internet, embedded into devices such as refrigerator, television etc so as to send and receive data (digital instructions). Such applications of technology results has revolutionized the way we use "human made things".

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4 years ago

The storage tanks are on a platform placed next to the building at the second-floor level. Based on the potential hazards, are t

LuckyWell [14K]

The should be no concerns about the space immediately below the tanks because there will be no damage to anything if any peril causes the tank to falls.

<h3>What are potential hazards?</h3>

Potential hazards refers to the active source for potential damage on a building, structure etc

It is important we know that the storage tanks are placed on a platform next to the building at the second-floor level but nothing beneath the tanks stands.

Hence, there should be no concerns about the space immediately below the tanks because there will be no damage to anything if any peril causes the tank to falls.

Read more about Potential hazards

<em>brainly.com/question/7310653</em>

5 0

2 years ago

# 45 You are driving on a two-lane highway and the vehicle behind you wants to pass. You should

hram777 [196]

Answer:

Signal the driver behind you when it is safe to pass by turning on your four-way emergency flashers.

Avoiding casualties is the top priority when driving, the other choices given do not put whether there are cars in the other lane into consideration, therefore making them incorrect. Signaling the driver when it is safe gives you the time and them the gateway to pass, making a nice interaction keep you both alive.

4 0

2 years ago

Water flows through a pipe of 100 mm at the rate of 0.9 m3 per minute at section A. It tapers to 50mm diameter at B, A being 1.5

pochemuha

Answer:

The velocities in points A and B are 1.9 and 7.63 m/s respectively. The Pressure at point B is 28 Kpa.

Explanation:

Assuming the fluid to be incompressible we can apply for the continuity equation for fluids:

$Aa.Va=Ab.Vb=Q$

Where A, V and Q are the areas, velocities and volume rate respectively. For section A and B the areas are:

$Aa=\frac{pi.Da^2}{4}= \frac{\pi.(0.1m)^2}{4}=7.85*10^{-3}\ m^3$

$Ab=\frac{pi.Db^2}{4}= \frac{\pi.(0.05m)^2}{4}=1.95*10^{-3}\ m^3$

Using the volume rate:

$Va=\frac{Q}{Aa}=\frac{0.9m^3}{7.85*10^{-3}\ m^3} = 1.9\ m/s$

$Vb = \frac{Q}{Ab}= \frac{0.9m^3}{1.96*10^{-3}\ m^3} = 7.63\ m/s$

Assuming no losses, the energy equation for fluids can be written as:

$Pa+\frac{1}{2}pa.Va^2+pa.g.za=Pb+\frac{1}{2}pb.Vb^2+pb.g.zb$

Here P, V, p, z and g represent the pressure, velocities, height and gravity acceleration. Considering the zero height level at point A and solving for Pb:

$Pb=Pa+\frac{1}{2}pa(Va^2-Vb^2)-pa.g.za$

Knowing the manometric pressure in point A of 70kPa, the height at point B of 1.5 meters, the density of water of 1000 kg/m^3 and the velocities calculated, the pressure at B results:

$Pb = 70000Pa+ \frac{1}{2}*1000\ \frac{kg}{m^3}*((1.9m/s)^2 - (7.63m/s)^2) - 1000\frac{kg}{m^3}*9,81\frac{m}{s^2}*1.5m$