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Sati [7]
4 years ago
7

Write a program that prompts the user to enter two characters and display the corresponding major and year status. The first cha

racter indicates the major. And the second character is a number character 1, 2, 3, 4, which indicates whether a student is a freshman, sophomore, junior, or senior. We consider only the following majors:
B (or b): Biology

C (or c): Computer Science

I (or i): Information Technology and Systems

Note that your program needs to let the user know if the major or year is invalid. Also, your program should be case-insensitive: your program should tell "Biology" either user type ‘b’ or ‘B’.

Engineering
1 answer:
lara31 [8.8K]4 years ago
7 0

Answer:

Following is attached the images for source code and the result according to given question.

I hope it will help you a lot!

Explanation:

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Because of the skin depth effect, the current at high frequency tends to flow at very low depth from radius. Then at high frequency the effective cross section of the wire is narrower than at DC.

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-1 1/6 divided by 2 1/3
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Answer:

-\frac{1}{2}

Explanation:

-1\frac{1}{6}:2\frac{1}{3}=\\  \\=-\frac{7}{6}:\frac{7}{3}\\  \\=-\frac{7}{6}*\frac{3}{7}\\  \\=-\frac{1}{2}

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4 years ago
A pin fin of uniform cross-sectional area is fabricated of an aluminum alloy (k = 160W m-1 K-1 ). The fin diameter is D = 4 mm,
disa [49]

Answer: (a) 36.18mm

(b) 23.52

Explanation: see attachment

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What is this spray pattern defect most likely caused by:
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6 0
2 years ago
10% A steel beam W18x76 spans 32 feet and is subjected to a Moment of 334 kips-ft. Find the load w on the beam. Determine the de
Lilit [14]

Answer:

w = 10.437 kips

deflection at 1/4 span  20.83\E ft

at mid span = 1.23\E ft

shear stress  7.3629 psi

Explanation:

area of cross section = 18*76

length of span = 32 ft

moment = 334 kips-ft

we know that

moment = load *eccentricity

334 = w * 32

w = 10.437 kips

deflection at 1/4 span

\delta = \frac{wa^2b^2}{3EI}

= \frac{10.4375*8^2 *24^2}{3E \frac{BD^3}{12}}

         =\frac{10.437 *8^2*24^2}{3E \frac{18*16^3}{12}}

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at mid span

\delta = \frac{wl^3}{48EI}

= \frac{10.43 *32^3}{48 *E*\frac{18*16^3}{12}}

\delta = 1.23\E ft

shear stress

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6 0
4 years ago
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