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Nutka1998 [239]
3 years ago
10

blank lines are used to see inside the product?blank lines are used to see inside the product what are they called ​

Engineering
1 answer:
mixer [17]3 years ago
7 0

Answer:

Do you have anymore information about this?

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As the amplitude decreases in a wave the energy increasesTrue or false
Montano1993 [528]

Answer:

True

Explanation:

To summarise, waves carry energy. The amount of energy they carry is related to their frequency and their amplitude. The higher the frequency, the more energy, and the higher the amplitude, the more energy.

7 0
3 years ago
Hydraulic fluid is primarily used to
Lerok [7]

Answer:

hydraulic flute is energy transfer medium in all hydraulic system this simple function is only achieved by flute that does not easily trap gases trapped gas and forming problems should bring a higher level of compressibility to a fluid than its usual at a point to support a very steep pass trekking system

8 0
3 years ago
The terms batten seam, standing seam, and flat seam all describe types of:
abruzzese [7]

Answer:

<em> (A) architectural sheet metal roofing</em>

Explanation:

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

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<em> </em><em>batten and standing seam</em>, and <em>flat seam all tells us that these are the types of</em> architectural sheet metal roofing.

5 0
3 years ago
Basic Question please help
Dvinal [7]
1(A)
2(B)
3(E)
4(C)
5(D)
6(B)
7(C)
7 0
3 years ago
In a Rankine cycle, superheated steam that enters the turbine at 1273.15 K and 1.8 MPa is then expanded to a vapor at 0.1 MPa. W
GrogVix [38]

Answer:

The shaft work generated per kilogram is -1.3 \frac{MJ}{kg}

Explanation:

Given:

Temperature T = 1273.15 K

Initial Pressure P_{1} = 1.8 MPa

Final pressure P_{2} = 0.1 MPa

From the table superheated,

h_{i} = 4635 \frac{K J}{Kg} and  h_{f} = 2706.54 \frac{K J}{Kg}

Work done by shaft is,

 W = h_{f} - h_{i}

 W = 2706.54 - 4635

 W = -1928.46 \frac{kJ}{kg}

But here efficiency is 0.56,

So work generated per kg is,

Work = 0.56 \times(- 1928.46)

Work = -1.3 \frac{MJ}{kg}

Therefore, the shaft work generated per kilogram is -1.3 \frac{MJ}{kg}

6 0
3 years ago
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