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katen-ka-za [31]
3 years ago
6

The top 3 most popular male names of 2017 are Oliver, Declan, and Henry according to babynames. Write a program that modifies th

e male_names set by removing a name and adding a different name. Sample output with inputs: 'Oliver' 'Atlas' { 'Atlas', 'Declan', 'Henry' } NOTE: Because sets are unordered, the order in which the names in male_names appear may differ from above.
Engineering
1 answer:
Flura [38]3 years ago
7 0

Answer:

The code is given below in Python with appropriate comments

Explanation:

# convert list to set

male_names = set(['Oliver','Declan','Henry'])

# get remove and add name from user

remove_name = input("Enter remove name: ")

add_name = input("Enter add name: ")

# remove name from set

male_names.remove(remove_name)

# add new name ij set

male_names.add(add_name)

# sort the set

a = sorted(male_names)

# print the set

print(a)

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Answer:0.1898 Pa/m

Explanation:

Given data

Diameter of Pipe\left ( D\right )=0.15m

Velocity of water in pipe\left ( V\right )=15cm/s

We know viscosity of water is\left (\mu\right )=8.90\times10^{-4}pa-s

Pressure drop is given by hagen poiseuille equation

\Delta P=\frac{128\mu \L Q}{\pi D^4}

We have asked pressure Drop per unit length i.e.

\frac{\Delta P}{L} =\frac{128\mu \ Q}{\pi D^4}

Substituting Values

\frac{\Delta P}{L}=\frac{128\times8.90\times10^{-4}\times\pi \times\left ( 0.15^{3}\right )}{\pi\times 4 \times\left ( 0.15^{2}\right )}

\frac{\Delta P}{L}=0.1898 Pa/m

4 0
3 years ago
A 2 in. diameter pipe supplying steam at 300°F is enclosed in a 1 ft square duct at 70°F. The outside of the duct is perfectly i
Shkiper50 [21]

Answer:

The value of heat transferred watt per foot length Q = 54.78 Watt per foot length.

Explanation:

Diameter of pipe = 2 in = 0.0508 m

Steam temperature T_{1} = 300 F  = 422.04 K

Duct temperature T_{2} = 70 F = 294.26 K

Emmisivity of surface 1 = 0.79

Emmisivity of surface 2 = 0.276

Net emmisivity of both surfaces ∈ = 0.25

Stefan volazman constant \sigma = 5.67 × 10^{-8} \frac{W}{m^{2} K^{4}  }

Heat transfer  per foot length is given by

Q = ∈ \sigma A ( T_{1}^{4} - T_{2} ^{4} ) ------ (1)

Put all the values in equation (1) , we get

Q = 0.25 × 5.67 × 10^{-8} × 3.14 × 0.0508 × 1 × ( 422.04^{4} - 294.26^{4} )

Q = 54.78 Watt per foot.

This is the value of heat transferred watt per foot length.

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Answer:

Explanation:

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