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motikmotik
3 years ago
13

Were you surprised by the “pie data”? Is it true for you, your family, and your friends? Why or why not?

Engineering
2 answers:
Ne4ueva [31]3 years ago
7 0

Answer: i was not suprised because, it is something all people need to know to gai  knowledege about it.

Sergeu [11.5K]3 years ago
6 0

Answer:

No because it is something to gain knowledge of peoples lives.

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Which is the required type of fire extinguisher for standard naval vessels
Bess [88]

Answer:

Mentioned below are the required types of fire extinguishers for standard naval vessels:

  1. Soda Acid Fire Extinguisher
  2. Water Extinguisher
  3. Foam Extinguisher – Chemical and Mechanical
  4. Carbon Dioxide Extinguisher
  5. Dry Powder Extinguisher

Explanation:

A fire extinguisher is a functioning fire insurance gadget used to douse or control little fires, regularly in crisis circumstances. It isn't planned for use on a wild fire, for example, one which has arrived at the roof, jeopardizes the client (i.e., no way out course, smoke, blast danger, and so on.), or in any case requires the mastery of a fire unit. Ordinarily, a fire extinguisher comprises of a hand-held barrel shaped weight vessel containing an operator that can be released to stifle a fire. Fire extinguishers made with non-round and hollow weight vessels likewise exist however are less normal.

A naval vessel is a military boat (or in some cases pontoon, contingent upon arrangement) utilized by a naval force. Naval boats are separated from non military personnel delivers by development and reason. By and large, naval boats are harm versatile and furnished with weapon frameworks, however combat hardware on troop transports is light or non-existent. Naval vessel is planned fundamentally for naval fighting are named warships, rather than help (assistant boats) or shipyard activities.

3 0
3 years ago
A bar of 75 mm diameter is reduced to 73mm by a cutting tool while cutting orthogonally. If the mean length of the cut chip is 7
barxatty [35]

Answer:

r=0.31

Ф=18.03°

Explanation:

Given that

Diameter of bar before cutting = 75 mm

Diameter of bar after cutting = 73 mm

Mean diameter of bar d= (75+73)/2=74 mm

Mean length of uncut chip = πd

Mean length of uncut chip = π x 74 =232.45 mm

So cutting ratio r

Cutting\ ratio=\dfrac{Mean\ length\ of cut\ chip}{Mean\ length\ of uncut\ chip}

r=\dfrac{73.5}{232.45}

  r=0.31

So the cutting ratio is 0.31.

As we know that shear angle given as

tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }

Now by putting the values

tan\phi =\dfrac{rcos\alpha }{1-rsin\alpha }

tan\phi =\dfrac{0.31cos15 }{1-0.31sin15 }\

  Ф=18.03°

So the shear angle is 18.03°.

4 0
3 years ago
11. As __and___ prices continued to rise in the late 1960’s and 70's, 4 and 6 cylinder engines began to make a comeback.
zvonat [6]

Answer:

Explanation:

whats the answeres two the question do they have a choise for you

5 0
1 year ago
A 1000 KVA three phase transformer has a secondary voltage of 208/120. What is the secondary full load amperage?
IceJOKER [234]

Answer:

The three phase full load secondary amperage is 2775.7 A

Explanation:

Following data is given,

S = Apparent Power = 1000 kVA

No. of phases = 3

Secondary Voltage: 208 V/120 V <em>(Here 208 V is three phase voltage and 120 V is single phase voltage) </em>

<em>Since,</em>

<em />

<em />V_{1ph} =\frac{ V_{3ph}}{\sqrt{3} }\\V_{1ph) = \frac{208}{\sqrt{3} }\\<em />

V_{1ph} = 120 V

The formula for apparent power in three phase system is given as:

S = \sqrt{3} VI

Where:

S = Apparent Power

V = Line Voltage

I = Line Current

In order to calculate the Current on Secondary Side, substituting values in above formula,

1000 kVA = \sqrt{3} * (208) * (I)\\1000 * 1000 = \sqrt{3} * (208) * (I)\\I = \frac{1000 * 1000}{\sqrt{3} * (208) }\\ I = 2775.7 A

 

4 0
3 years ago
simple Brayton cycle using air as the working fluid has a pressure ratio of 10. The minimum and maximum temperatures in the cycl
Irina18 [472]

Answer:

a) 764.45K

b) 210.48 kJ/kg

c) 30.14%

Explanation:

pressure ratio = 10

minimum temperature = 295 k

maximum temperature = 1240 k

isentropic efficiency for compressor = 83%

Isentropic efficiency for turbine = 87%

<u>a) Air temperature at turbine exit </u>

we can achieve this by interpolating for enthalpy

h4 = 783.05 kJ/kg ( calculated in the background ) at state 4 using Table A-17  for  Ideal gas properties of air

T4 ( temperature at Turbine exit ) = 760 + ( 780 - 760 ) (\frac{783.05-778.18}{800.13-778.18} ) = 764.45K

<u>b) The net work output </u>

first we determine the actual work input to compressor

Wc = h2 - h1  ( calculated values )

     = 626.57 - 295.17 =  331.4 kJ/kg

next determine the actual work done by Turbine

Wt = h3 - h4  ( calculated values )

     = 1324.93 - 783.05 = 541.88 kJ/kg

finally determine the network output of the cycle

Wnet = Wt - Wc

         = 541.88 - 331.4  = 210.48 kJ/kg

<u>c) determine thermal efficiency </u>

лth = Wnet / qin  ------ ( 1 )

where ; qin = h3 - h2

<em>equation 1 becomes </em>

лth = Wnet / ( h3 - h2 )

      = 210.48 / ( 1324.93 - 626.57 )

      = 0.3014  =  30.14%

6 0
3 years ago
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