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antoniya [11.8K]

2 years ago

7

If extension done in a stork position, during backward bending in a standing position of a special test for lumbar spine, produc

es pain in the lumbar or sacral regions of the spine, it may indicate _____.

1 answer:

Klio2033 [76]2 years ago

5 0

The answer is 7 because I just took the test!

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The underground cafe has an operating cash flow of $187,000 and a cash flow to creditors of $71,400 for the past year. During th

Serggg [28]

Answer:

cash flow to stockholders = $39,700

Explanation:

Operating cash flow = $187,000

cash flow to creditors = $71,400

Net working capital = $28,000

Net capital spending = $47,900

Cash flow to stockholders = ?

CFF = operating cash flow - net working capital - net capital spending

CFF = $187,000 - $28,000 - $47,900 = $111,100

CFF = cash flow to creditors + cash flow to stockholders

cash flow to stockholders = CFF - cash flow to creditors

cash flow to stockholders = $111,100 - $71,400 = $39,700

Hence $39,700 is the amount of the cash flow to stockholders for the last year.

3 0

3 years ago

Explain when it is appropriate to use Tier II?

garik1379 [7]

????? what do u mean tier 2

3 0

3 years ago

What happens to the amperage draw of a condensing unit on a split AC system if the liquid line is restricted

cupoosta [38]

Answer:

The amperage draw of the condensing unit will be low.

Explanation:

A condensing unit is made up of a compressor and condenser, while an evaporating unit is made up of an evaporator coil.

A split AC system is a type of air conditioner system that has a condensing unit which is placed separately from the evaporative coil unit. Then the two units are connected to each other via a copper tube containing refrigerants.

The liquid line connects the condenser to the evaporator, and if this liquid line is restricted, the amp consumed by the condensing unit will be low.

6 0

3 years ago

Air flows through a convergent-divergent duct with an inlet area of 5 cm² and an exit area of 3.8 cm². At the inlet section, the

Luda [366]

Answer:

The mass flow rate is 0.27 kg/s

The exit velocity is 76.1 m/s

The exit pressure is 695 KPa

Explanation:

Assuming the flow to be steady state and the behavior of air as an ideal gas.

The mass flow rate of the air is given as:

Mass Flow Rate = ρ x A1 x V1

where,

ρ = density of air

A1 = inlet area = 3.8 cm² = 3.8 x 10^-4 m²

V1 = inlet velocity = 100 m/s

For density using general gas equation:

PV = nRT

PV = (m/M)RT

PM/RT = ρ

ρ = (680000 N/m²)(0.02897 kg/mol)/(8.314 J/mol.k)(60 + 273)k

ρ = 7.11 kg/m³

Therefore,

Mass Flow Rate = (7.11 kg/m³)(3.8 x 10^-4 m²)(100 m/s)

<u>Mass Flow Rate = 0.27 kg/s = 270 g/s</u>

Now, for steady flow, the mass flow rate remains constant throughout the flow. Hence, flow rate at inlet will be equal to the flow rate at outlet:

Mass Flow Rate = ρ x A2 x V2

where,

ρ = density of air = 7.11 kg/m³ (Assuming in-compressible flow)

A2 = exit area = 5 cm² = 5 x 10^-4 m²

V2 = exit velocity = ?

Therefore:

0.27 kg/s = (7.11 kg/m³)(5 x 10^-4 m²) V2

<u>V2 = 76.1 m/s</u>

Now, for exit pressure, we use Bernoulli's equation between inlet and exit, using subscript 1 for inlet and 2 for exit:

P1 + (1/2) ρ V1² + ρ g h1 = P2 + (1/2) ρ V2² + ρ g h2

Since, both inlet and exit are at same temperature.

Therefore, h1 = h2, and those terms will cancel out.

P1 + (1/2) ρ V1² = P2 + (1/2) ρ V2²

P2 = P1 + (1/2) ρ V1² - (1/2) ρ V2²

P2 = P1 + (1/2) ρ (V1² - V2²)

P2 = 680000 Pa + (0.5)(7.11 kg/m³)[(100m/s)² - (76.1 m/s)²]

P2 = 680000 Pa + 14962.25 Pa

<u>P2 = 694962.25 Pa = 695 KPa</u>

4 0

3 years ago

The ingredient weights for making 1 yd (cyd) of concrete by assuming aggregates in SSD state are given below. The volume of air

Pachacha [2.7K]

Answer:

Explanation:

Ans) Given batch weight of each component :

Cement = 700 lb

Water = 315 lb

Coarse aggregate = 1575 lb

Fine aggregate = 1100 lb

Part 1) Amount of water = 328.5 lb

Amount of water is needed to be increased if the aggregates has absorption capacity, To maintain constant water cement ratio, the mixing water is increased because some of the water is absorbed by aggregates.

Amount of water absorbed = 328.5 lb - 315 lb = 13.5 lb

Total amount of aggregates = 1575 + 1100 = 2675 lb

=> % Absorption capacity = 13.5 x 100 / 2675 = 0.5 %

Hence, new amount of Coarse aggregate = (1 - 0.005) x 1575 lb = 1567.125 lb

New amount of fine aggregate = (1 - 0.005) x 1100 = 1094.5 lb

Since, water cement ratio is maintained constant , amount of cement remains unchanged

=> Volume of water = 328.5 / 62.4 = 5.26 ft3

=> Volume of cement = 700 / (3.15 x 62.4) = 3.56 ft3

=> Volume of coarse aggregate = 1567.125 / (2.4 x 62.4) = 10.46 ft3

=> Volume of fine aggregate = 1100 / (2.4 x 62.4) = 7.34 ft3

Volume of air = 2% = 0.02 x 27 = 0.54 ft3

Total concrete volume = 5.26 + 3.56 + 10.46 + 7.34 + 0.54 \approx 27 ft3 = 1 yd3

Hence, calculated amount of each component is correct

Part 2) We know, minus sign indicated that the aggregate will absorb some moisture from concrete, hence mixing water amount needed to be corrected .

=> Amount of water absorbed by coarse aggregate = 0.01 x 1567.125 lb = 15.67 lb

=> Amount of water absorbed by fine aggregate = 0.02 x 1094.50 lb = 21.89 lb

Total amount of water absorbed = 15.67 + 21.89 = 37.56 lb

To maintain same water cement ratio, amount of mixing water is needed to be increased

=> Corrected amount of mixing water = 328.5 lb + 37.56 lb = 366 lb

=> Corrected amount of coarse aggregate = (1 - 0.01) x 1567.125 = 1551.45 lb

=> Corrected amount of fine aggregate = (1 - 0.02) x 1094.5 = 1072.6 lb

Part 3) We know,

Unit weight = Sum of weight of each material / Total volume

=> Sum of weight = 366 + 700 + 1551.45 + 1072.6 = 3690.05 lb

Total volume = 1 yd3 or 27 ft3

=> Expected Unit Weight = 3690.05 lb / 27 ft3 = 136.67 lb/ft3

Also, Concrete Yield = Weight of all components / Unit weight of concrete

=> Yield = 3690.05 / 136.67 = 27 ft3 or 1 yd3

4 0

4 years ago