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3 years ago

I NEED HELP ASAP!!

Engineering

2 answers:

Annette [7]3 years ago

8 0

Answer:

7.The number of turns should be such that the induced counter electromotive force can maintain the balance of the applied voltage.

I think that is the answer

shusha [124]3 years ago

3 0

7.The number of turns should be such that the induced counter electromotive force can maintain the balance of the applied voltage. It should be that answer :)

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Water enters a tank from two pipes, one with a flow rate of 0.3 kg/s and the other with a flow rate of 0.1 kg / s. The tank has

stiks02 [169]

Answer:

total amount of water after 2 min will be 84.4 kg/s

Explanation:

Given data:

one tank inflow = 0.1 kg/s

2nd tank inflow = 0.3 kg/s

3rd tank outflow = 0.03 kg/s

Total net inflow in tank is = 0.3 +0.1 =0.4 kg/s

From third point, outflow is 0.03 kg/s

Therefore, resultant in- flow = 0.4 - 0.03

Resultant inflow is = 0.37 kg/s

Tank has initially 40 kg water

In 2 min ( 2*60 sec), total inflow in tank is 0.37*60*2 = 44.4 kg

So, total amount of water after 2 min will be = 40+44.4 = 84.4 kg

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3 years ago

Question 7.1: Two possible overhead valve combustion chambers are being considered – the first has two valves; the second has fo

AleksandrR [38]

Answer:

1) The adoption of the second design we can see that the total valve perimeter is increased by 60.8%

2) Increase in flow are : 29%

3) Additional benefits in using 4 valves per cylinder:

a)For the purpose of controlling the combustion process, the inlet valves will give more flexibility

b) There is a larger valve throat areas for the flow of gas

Explanation:

1) Perimeter of the first possible overhead valve combustion chamber with two valves:

P₂ = πd = π × 23 = 72.26mm

Perimeter of the second possible overhead valve combustion chamber with four valves:

P₄ = π2d = π × 18.5 × 2 = 116.24 mm

If second design is adopted, percentage increase = ((P₄ - P₂)/P₂)×100

= ((116.24 - 72.26)/72.26)×100 = 0.6086 ×100 = 60.86%

Therefore, the total valve perimeter is shown to have increased by 60.8%

2) Formula for flow Area (A) = P × L = πkd²

Area of the first possible overhead valve combustion chamber with two valves: A₂ = πkd² = πk(23)² = 1662k mm²

Area of the first possible overhead valve combustion chamber with four valves: A₄ = πkd² = 2πk(18.5)² = 2150k mm²

The percentage increase in flow area: ((A₄ - A₂)/A₄)×100 = ((2150 - 1662)/2150)×100 = 29%

3) The additional benefits of using are:

a) For the purpose of controlling the combustion process, the inlet valves will give more flexibility

b) There is a larger valve throat areas for the flow of gas

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3 years ago

Why is sssniperwolf so fine

RSB [31]

Bc she’s just a baddie

5 0

3 years ago

Read 2 more answers

The terms batten seam, standing seam, and flat seam all describe types of:

abruzzese [7]

Answer:

 (A) architectural sheet metal roofing

Explanation:

By the name itself we can judge that the 'Architectural sheet metal roofing' is a kind of metal roofing.

And these type of metal roofing is primarily used for small and big houses, small buildings and as well as in a building that is for commercial use they can be totally flat as well as little bit sloped.

And the words similarly like batten and standing seam, and flat seam all tells us that these are the types of architectural sheet metal roofing.

5 0

3 years ago

Steam enters a radiator at 16 psia and 0.97 quality. The steam flows through the radiator, is con- densed, and leaves as liquid

AnnZ [28]

Answer:

5.328Ibm/hr

Explanation:

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties such as pressure and temperature.

for this case we can define the following equation for mass flow using the first law of thermodynamics

$m=\frac{Q}{h1-h2}$

where

Q=capacity of the radiator =5000btu/hr

m = mass flow

then using thermodynamic tables we found entalpy in state 1 and 2

h1(x=0.97, p=16psia)=1123btu/lbm

h2(x=0, p=16psia)=184.5btu/lbm

solving

$Q=\frac{5000}{1123-184.5} =5.328Ibm/hr$

3 0

3 years ago