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Consider a refrigerator that consumes 320 W of electric power when it is running. If the refrigerator runs only one quarterof th

ryzh [129]

Answer:

$5.184

Explanation:

The cost can be calculated using the formula: $Cost = Load \ factor \times Number \ of \ hours \ \\M_{month} = M_{units} \times W\\$

Before using this, we require the following conversions:

320 W → kW:

$\frac {320}{1000} = 0.32$

30 Days → Hours:

$30 \times 24 = 720$

Using the above stated formula:

$M_{month} = 0.09 \times 0.32 \times \frac{1}{4} \times 720 = 5.184$

4 0

3 years ago

Rear axles are usually lubricated by the same gear oil that lubricates the differential. True or false

KIM [24]

Answer:

true

Explanation:

8 0

2 years ago

Pipe Diameter and Reynolds Number. An oil is being pumped inside a 10.0-mm-diameter pipe at a Reynolds number of 2100. The oil d

alexdok [17]

Answer:

The velocity in the pipe is 5.16m/s. The pipe diameter for the second fluid should be 6.6 mm.

Explanation:

Here the first think you have to consider is the definition of the Reynolds number ( $Re$ ) for flows in pipes. Rugly speaking, the Reynolds number is an adimensonal parameter to know if the fliud flow is in laminar or turbulent regime. The equation to calculate this number is:

$Re=\frac{\rho v D}{\mu}$

where $\rho$ is the density of the fluid, $\mu$ is the viscosity, D is the pipe diameter and v is the velocity of the fluid.

Now, we know that Re=2100. So the velocity is:

$v=\frac{Re*\mu}{\rho*D} =\frac{2100*2.1x10^{-2}Pa*s }{855kg/m^3*0.01m} =5.16m/s$

For the second fluid, we want to keep the Re=2100 and v=5.16m/s. Therefore, using the equation of Reynolds number the diameter is:

$D=\frac{Re*\mu}{\rho*v} =\frac{2100*1.5x10^{-2}Pa*s}{925kg/m^3*5.16m/s}=6.6 mm$

8 0

3 years ago

You guys are amazing :D

Sloan [31]

Ik i am thank you tho xoxo

3 0

3 years ago

Read 2 more answers

A 10.2 mm diameter steel circular rod is subjected to a tensile load that reduces its cross- sectional area to 52.7 mm^2. Determ

VMariaS [17]

Answer:

The percentage ductility is 35.5%.

Explanation:

Ductility is the ability of being deform under applied load. Ductility can measure by percentage elongation and percentage reduction in area. Here, percentage reduction in area method is taken to measure the ductility.

Step1

Given:

Diameter of shaft is 10.2 mm.

Final area of the shaft is 52.7 mm².

Calculation:

Step2

Initial area is calculated as follows:

$A=\frac{\pi d^{2}}{4}$

$A=\frac{\pi\times(10.2)^{2}}{4}$

A = 81.713 mm².

Step3

Percentage ductility is calculated as follows:

$D=\frac{A_{i}-A_{f}}{A_{i}}\times100$

$D=\frac{81.713-52.7}{81.713}\times100$

D = 35.5%.

Thus, the percentage ductility is 35.5%.

5 0

3 years ago