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Vanyuwa [196]
3 years ago
11

T

Engineering
1 answer:
solniwko [45]3 years ago
3 0

Answer:

Common Uses: Boxwood is well-suited for carving and turning, and the tree's diminutive size restricts it to smaller projects. Some common uses for Boxwood include: carvings, chess pieces, musical instruments (flutes, recorders, woodwinds, etc.), rulers, handles, turned objects, and other small specialty items.If you want a small, compact, low-growing shrub to form a hedge that serves as an accent or border along your walkway, fence line or planting beds, dwarf boxwood varieties are the best pick. The "Dwarf English" boxwood (Buxus sempervirens “Suffruticosa”) creates a border hedge approximately 1 to 2 feet in height.

Explanation:

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Which option identifies the type of engineering technician most likely to be involved in the following scenario?
aleksandrvk [35]

Answer:

I believe the answer to your question would be A.

Explanation:

Electrical engineering technician

6 0
2 years ago
11. Technicians A and B are discussing
algol [13]

Answer:

C. Neither Technician A nor B

Explanation:

Just took the test

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3 years ago
16. Driverless cars have already , and they look so cool.
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Hope it helps!
4 0
3 years ago
Read 2 more answers
Suppose you have two arrays: Arr1 and Arr2. Arr1 will be sorted values. For each element v in Arr2, you need to write a pseudo c
brilliants [131]

Answer:

The algorithm is as follows:

1. Declare Arr1 and Arr2

2. Get Input for Arr1 and Arr2

3. Initialize count to 0

4. For i in Arr2

4.1 For j in Arr1:

4.1.1 If i > j Then

4.1.1.1 count = count + 1

4.2 End j loop

4.3 Print count

4.4 count = 0

4.5 End i loop

5. End

Explanation:

This declares both arrays

1. Declare Arr1 and Arr2

This gets input for both arrays

2. Get Input for Arr1 and Arr2

This initializes count to 0

3. Initialize count to 0

This iterates through Arr2

4. For i in Arr2

This iterates through Arr1 (An inner loop)

4.1 For j in Arr1:

This checks if current element is greater than current element in Arr1

4.1.1 If i > j Then

If yes, count is incremented by 1

4.1.1.1 count = count + 1

This ends the inner loop

4.2 End j loop

Print count and set count to 0

<em>4.3 Print count</em>

<em>4.4 count = 0</em>

End the outer loop

4.5 End i loop

End the algorithm

5. End

6 0
3 years ago
What is the maximal coefficient of performance of a refrigerator which cools down 10 kg of water (and then ice) to -6C. Upper he
inysia [295]

Given:

Temperature of water, T_{1} = -6^{\circ}C =273 +(-6) =267 K

Temperature surrounding refrigerator, T_{2} = 21^{\circ}C =273 + 21 =294 K

Specific heat given for water, C_{w} = 4.19 KJ/kg/K

Specific heat given for ice, C_{ice} = 2.1 KJ/kg/K

Latent heat of fusion,  L_{fusion} = 335KJ/kg

Solution:

Coefficient of Performance (COP) for refrigerator is given by:

Max COP_{refrigerator} = \frac{T_{2}}{T_{2} - T_{1}}

= \frac{267}{294 - 267} = 9.89

Coefficient of Performance (COP) for heat pump is given by:

Max COP_{heat pump} = \frac{T_{1}}{T_{2} - T_{1}}\frac{294}{294 - 267} = 10.89

6 0
3 years ago
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