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zavuch27 [327]
3 years ago
9

The following program includes fictional sets of the top 10 male and female baby names for the current year. Write a program tha

t creates: A set all_names that contains all of the top 10 male and all of the top 10 female names. A set neutral_names that contains only names found in both male_names and female_names. A set specific_names that contains only gender specific names. Sample output for all_names: {'Michael', 'Henry', 'Jayden', 'Bailey', 'Lucas', 'Chuck', 'Aiden', 'Khloe', 'Elizabeth', 'Maria', 'Veronica', 'Meghan', 'John', 'Samuel', 'Britney', 'Charlie', 'Kim'}

Engineering
2 answers:
yuradex [85]3 years ago
4 0

Answer:

Please see attachment

Explanation:

Please see attachment

otez555 [7]3 years ago
3 0

Answer:

Define Variables and Use List methods to do the following

Explanation:

#<em>Conjoins two lists together</em>

all_names = male_names.union(female_names)

#<em>Finds the names that appear in both lists, just returns those</em>

neutral_names = male_names.intersection(female_names)

#<em>Returns names that are NOT in both lists</em>

specific_names = male_names.symmetric_difference(female_names)

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A 240 V, 60 Hz squirrel-cage induction motor has a full-load slip of 0.02 and a full-load speed of 1764 rpm. The winding resista
earnstyle [38]

Answer:

full load current = 7.891151 A

so correct option is b. Ir= 7.89

Explanation:

given data

Energy E = V = 240 V

frequency =  60 Hz

full-load slip = 0.02

full-load speed = 1764 rpm

winding resistance = 0.6 Ω

winding reactance = 5 Ω

to find out

rotor current at full-load

solution

we will apply here full load current formula that is express as

full load current = \frac{S*E}{\sqrt{R^2+ (S*X)^2}}    ...................1

here S is full-load slip and E is energy given and R is winding resistance and    X is winding reactance

put here value we get

full load current = \frac{0.02*240}{\sqrt{0.6^2+ (0.02*5)^2}}  

full load current = 7.891151 A

so correct option is b. Ir= 7.89

6 0
3 years ago
A satellite orbits the Earth every 2 hours at an average distance from the Earth's centre of 8000km. (i) What is the average ang
AlexFokin [52]

Answer:

i)ω=3600 rad/s

ii)V=7059.44 m/s

iii)F=1245.8 N

Explanation:

i)

We know that angular speed given as

\omega =\dfrac{d\theta}{dt}

We know that for one revolution

θ=2π

Given that time t= 2 hr

So

ω=θ/t

ω=2π/2 = π rad/hr

ω=3600 rad/s

ii)

Average speed V

V=\sqrt{\dfrac{GM}{R}}

Where M is the mass of earth.

R is the distance

G is the constant.

Now by putting the values

V=\sqrt{\dfrac{GM}{R}}

V=\sqrt{\dfrac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{8000\times 10^3}}

V=7059.44 m/s

iii)

We know that centripetal fore given as

F=\dfrac{mV^2}{R}

Here given that m= 200 kg

R= 8000 km

so now by putting the values

F=\dfrac{mV^2}{R}

F=\dfrac{200\times 7059.44^2}{8000\times 10^3}

F=1245.8 N

3 0
3 years ago
Based on bonding theory, explain why coefficient of expansion increases when you consider ceramics, metals and polymers
Mrrafil [7]

Answer:

The ratio that a material expands in accordance with changes in temperature is called the coefficient of thermal expansion. Because Fine Ceramics possess low coefficients of thermal expansion, their distortion values, with respect to changes in temperature, are low. The coefficients of thermal expansion depend on the bond strength between the atoms that make up the materials.

Explanation:

:>

3 0
4 years ago
A sleeve made of SAE 4150 annealed steel has a nominal inside diameter of 3.0 inches and an outside diameter of 4.0 inches. It i
irga5000 [103]

Answer:

Class of fit:

Interference (Medium Drive Force Fits constitute a special type of Interference Fits and these are the tightest fits where accuracy is important).

Here minimum shaft diameter will be greater than the maximum hole diameter.

Medium Drive Force Fits are FN 2 Fits.

As per standard ANSI B4.1 :

Desired Tolerance: FN 2

Tolerance TZone: H7S6

Max Shaft Diameter: 3.0029

Min Shaft Diameter: 3.0022

Max Hole Diameter:3.0012

Min Hole Diameter: 3.0000

Max Interference: 0.0029

Min Interference: 0.0010

Stress in the shaft and sleeve can be considered as the compressive stress which can be determined using load/interference area.

Design is acceptable If compressive stress induced due to selected dimensions and load is less than compressive strength of the material.

Explanation:

4 0
3 years ago
Read 2 more answers
Automobiles must be able to sustain a frontal impact. Specifically, the design must allow low speed impacts with little damage,
Naya [18.7K]

Answer:

Explanation:

The concept of Hooke's law was applied as it relates to deformation.

The detailed steps and appropriate substitution is as shown in the attached file.

8 0
3 years ago
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