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o-na [289]
3 years ago
8

2.4 kg of nitrogen at an initial state of 285K and 150 kPa is compressed slowly in an isothermal process to a final pressure of

600 kPa. Determine the work.
Engineering
1 answer:
Neko [114]3 years ago
8 0

Answer:

W=-280.67 KJ

Explanation:

Given that

Initial pressure = 150 KPa

Final pressure = 600 KPa

Temperature T= 285 K

Mass m=2.4 Kg

We know that ,work in isothermal process given as

W=mRT\ ln\dfrac{P_1}{P_2}

Gas constant for nitrogen gas R=0.296 KJ/kgK

Now by putting the values

W=mRT\ ln\dfrac{P_1}{P_2}

W=2.4\times 0.296\times 285\ ln\dfrac{150}{600}

W=-280.67 KJ

Negative sign indicates that it is compression process and work is done on the gas.

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Cold forging makes metal more workable than hot forging.<br> True<br> False
Whitepunk [10]

Answer:

I don't really know but i have some info for you...

Explanation:

The cold forging manufacturing process increases the strength of a metal through strain hardening at a room temperature. On the contrary the hot forging manufacturing process keeps materials from strain hardening at high temperature, which results in optimum yield strength, low hardness and high ductility.

7 0
3 years ago
Read 2 more answers
The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8 What is the probability of observing
Viktor [21]

Answer:

0.14% probability of observing more than 4 errors in the carpet

Explanation:

When we only have the mean, we use the Poisson distribution.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8.

This means that \mu = 0.8

What is the probability of observing more than 4 errors in the carpet

Either we observe 4 or less errors, or we observe more than 4. The sum of the probabilities of these outcomes is 1. So

P(X \leq 4) + P(X > 4) = 1

We want P(X > 4). Then

P(X > 4) = 1 - P(X \leq 4)

In which

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-0.8}*(0.8)^{0}}{(0)!} = 0.4493

P(X = 1) = \frac{e^{-0.8}*(0.8)^{1}}{(1)!} = 0.3595

P(X = 2) = \frac{e^{-0.8}*(0.8)^{2}}{(2)!} = 0.1438

P(X = 3) = \frac{e^{-0.8}*(0.8)^{3}}{(3)!} = 0.0383

P(X = 4) = \frac{e^{-0.8}*(0.8)^{4}}{(4)!} = 0.0077

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.4493 + 0.3595 + 0.1438 + 0.0383 + 0.0077 = 0.9986

P(X > 4) = 1 - P(X \leq 4) = 1 - 0.9986 = 0.0014

0.14% probability of observing more than 4 errors in the carpet

5 0
3 years ago
Question Set 22.1 Using the count method, find the number of occurrences of the character 's' in the string 'mississippi'.2.2 In
Gnom [1K]

Answer:

# Program is written in python

# 22.1 Using the count method, find the number of occurrences of the character 's' in the string 'mississippi'.

# initializing string

Stringtocheck = "mississippi"

# using count() to get count of s

counter = Stringtocheck.count('s')

# printing result

print ("Count of s is : " + str(counter))

# 2.2 In the string 'mississippi', replace all occurrences of the substring 'iss' with 'ox

# Here, we'll make use of replace() method

# Prints the string by replacing iss by ox

print(Stringtocheck.replace("iss", "ox"))

#2.3 Find the index of the first occurrence of 'p' in 'mississippi'

# declare substring

substring = 'p'

# Find index

index = Stringtocheck.find(substring)

# Print index

print(index)

# End of program

8 0
2 years ago
What is the volume of the rectangular prism shown.
Likurg_2 [28]
It would be 72cm bc u need to add up all the line in the back to
7 0
3 years ago
Please help me with this. Plzzz.
Drupady [299]

Answer:

450,000m = 450km = 4.5E5

32,600,000W = 32.6MW = 3.26E7

59,700,000,000cal = 59.7Gcal = 5.97E10

0.000000083s = 83ns = 8.3E-8

35,000Ω = 35kΩ = 3.5E4

Explanation:

Giga   = 1,000,000,000

Mega = 1,000,000

kilo     = 1,000

unit    = 1

deci   = .1

centi  = .01

milli    = .001

micro = .000001

nano = .0000000001

pico  = .000000000001

You should be able to look at these and convert between them in seconds if you want to pursue anything in engineering.

7 0
3 years ago
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