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Sati [7]
3 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]3 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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An electrical heater is a form of sensible heating process, and heats 0.1m/s of air from 15°C and 80% RH to 50°C? The barometric
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Answer:

The heater load =35 KJ/kg

Explanation:

Given that

At initial condition

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At final condition

Temperature= 50°C

We know that in sensible heating process humidity ratio remain constant.

Now from chart

At temperature= 15°C and RH=80%

h_1=38 \frac{KJ}{kg},v=0.8 \frac{m^3}{kg}

At  temperature= 50°C

h_2=73 \frac{KJ}{kg}

So\ the\ heater\ load =h_2-h_1

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2 years ago
What is the metal removal rate when a 2 in-diameter hole 3.5 in deep is drilled in 1020 steel at cutting speed of 120 fpm with a
Studentka2010 [4]

Answer:

a) the metal removal rate is 14.4 in³/min

b) the cutting time is 0.98 min

Explanation:

Given the data from the question

first we find the rpm for the spindle of the drilling tool, using the equation

Ns = 12V/πD

V is the cutting speed(120 fpm) and D is the diameter of the hole( 2 in)

so we substitute

Ns = 12 × 120 / π2

Ns = 1440 / 6.2831

Ns = 229.18 rmp

Now we find the metal removal rate using the equation

MRR = (πD²/4) Fr × Ns

Fr is the feed rate( 0.02 ipr ),

so we substitute

MRR = ((π × 2²)/4) × 0.02 × 229.18

MRR = 14.3998 ≈ 14.4 in³/min

Therefore the metal removal rate is 14.4 in³/min

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A = D/2

A = 2/2

= 1 in

now find the time required to drill the hole

Tm = (L + A) / (Fr × Ns)

Lis the the depth of the hole( 3.5 in)

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Tm = (3.5 + 1) / (0.02 × 229.18  )

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