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timurjin [86]
3 years ago
7

Write a program that asks the user to enter a list of numbers. The program should take the list of numbers and add only those nu

mbers between 0 and 100 to a new list. It should then print the contents of the new list. Running the program should look something like this:
Please enter a list of numbers: 10.5 -8 105 76 83.2 206

The numbers between 0 and 100 are: 10.5 76.0 83.2
Engineering
1 answer:
pshichka [43]3 years ago
4 0

In python 3.8

nums = input("Please enter a list of numbers: ").split()

new_nums = [x for x in nums if 0 < float(x) < 100]

print("The numbers between 0 and 100 are: " + " ".join(new_nums))

When you said numbers between 0 and 100, I didn't know if that was inclusive or exclusive so I made it exclusive. I hope this helps!

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Kobotan [32]

Answer:

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An 18-in.-long titanium alloy rod is subjected to a tensile load of 24,000 lb. If the allowable tensile stress is 60 ksi and the
Vilka [71]

Answer:

Required Diameter = 302.65 inches

Explanation:

We are given;

Allowable tensile stress = 60 ksi

Weight of tensile load = 24,000 lb

Elongation = 0.05 in

Original length = 18 in

We'll need to check the diameters under stress and strain.

Now, we know that the formula for stress is;

Stress = Force/Area

Thus,

Area = Force/stress

So for this stress, area required is;

A_req = 24000/60 = 4000 in²

So let's find the required diameter here.

Area = πd²/4

So, 4000 = πd²/4

(4000 x 4)/π = d²

d² = 5092.96

Required diameter here is;

d = √5092.96

d = 71.36 in

For Strain;

Formula for strain is;

Strain = stress/E

We are given E = 120 ksi

stress = P/A = 24,000/A

strain = elongation/original length = 0.05/18 = 0.00278

Thus;

0.00278 = P/(A•E)

0.00278 = 24000/(120 x A)

Making A the subject to obtain;

A = 24000/(120 x 0.00278)

A_required = 71942 in²

Area = πd²/4

So, 71942 = πd²/4

(71942 x 4)/π = d²

d² = 91599.4

Required diameter here is;

d = √91599.4

d = 302.65 in

The larger diameter is 302.65 inchesand it's therefore the required one.

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3 years ago
Read 2 more answers
It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d
denis23 [38]

The exit temperature is 586.18K and  compressor input power is 14973.53kW

Data;

  • Mass = 50kg/s
  • T = 288.2K
  • P1 = 1atm
  • P2 = 12 atm

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\

Let's substitute the values into the formula

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW

The compressor input power is 14973.53kW

Learn more on exit temperature and compressor input power here;

brainly.com/question/16699941

brainly.com/question/10121263

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