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2 years ago

Which apparatus is included in a task force?

Engineering

1 answer:

Yakvenalex [24]2 years ago

7 0

Answer:

I am Sachin this answer is fire engine is your name is this

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Answer back to question for la ,lot points

Lerok [7]

Answer:

yes

Explanation:

yes

5 0

2 years ago

I am trying to make a vacuum cannon but all I can use to get out the air is a speed pump to give air to bicycles. I need to make

Mrrafil [7]

We can actually deduce here that making a airtight seal will take different format. You can:

Use an epoxy-resin to create an airtight seal
Create a glass-metal airtight seal
Make a ceramic-metal airtight seal.

<h3>What is an airtight seal?</h3>

An airtight seal is actually known to be a seal or sealing that doesn't permit air or gas to pass through. Airtight seal are usually known as hermetic seal. They are usually applied to airtight glass containers but the advancement in technology has helped to broaden the materials.

We can see that epoxy-resin can used to create an airtight seal. They create airtight seals to copper, plastics, stainless steels, etc.

When making glass-metal airtight seal, the metal should compress round the solidified glass when it cools.

Learn more about airtight seal on brainly.com/question/14977167

#SPJ1

6 0

1 year ago

An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500

Assoli18 [71]

Answer: The exit temperature of the gas in deg C is $32^{o}C$ .

Explanation:

The given data is as follows.

$C_{p}$ = 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

$P_{1}$ = 100 kPa, $V_{1} = 15 m^{3}/s$

$T_{1} = 27^{o}C = (27 + 273) K = 300 K$

We know that for an ideal gas the mass flow rate will be calculated as follows.

$P_{1}V_{1} = mRT_{1}$

or, m = $\frac{P_{1}V_{1}}{RT_{1}}$

= $\frac{100 \times 15}{0.5 \times 300}$

= 10 kg/s

Now, according to the steady flow energy equation:

$mh_{1} + Q = mh_{2} + W$

$h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}$

$C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}$

$(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}$

$(T_{2} - T_{1})$ = 5 K

$T_{2}$ = 5 K + 300 K

$T_{2}$ = 305 K

= (305 K - 273 K)

= $32^{o}C$

Therefore, we can conclude that the exit temperature of the gas in deg C is $32^{o}C$ .

8 0

3 years ago

Answer my question I will mark brainliest

Iteru [2.4K]

Answer:

150

Explanation:

Mark me Brainliest

6 0

2 years ago

Read 2 more answers

2. A well of 0.1 m radius is installed in the aquifer of the preceding exercise and is pumped at a rate averaging 80 liter/min.

hodyreva [135]

Question:

The question is not complete. See the complete question and the answer below.

A well that pumps at a constant rate of 0.5m3/s fully penetrates a confined aquifer of 34 m thickness. After a long period of pumping, near steady state conditions, the measured drawdowns at two observation wells 50m and 100m from the pumping well are 0.9 and 0.4 m respectively. (a) Calculate the hydraulic conductivity and transmissivity of the aquifer (b) estimate the radius of influence of the pumping well, and (c) calculate the expected drawdown in the pumping well if the radius of the well is 0.4m.

Answer:

T = 0.11029m²/sec

Radius of influence = 93.304m

expected drawdown = 3.9336m

Explanation:

See the attached file for the explanation.

8 0

3 years ago