Compute the weight fraction of graphite, in a 3 wt% C Ferritic Gray cast iron, assuming that all the carbon exists as the graphite phase. Assume densities of 7.9 and 2.3 g/cm3 for ferrite and graphite, respectively.

Answers

Answer 1

Answer:

9.60%

Explanation:

Computation for the weight fraction of graphite

First step

Computation for the mass fraction for Wa using this formula

Wa=Cg-Co/Cg-Ca

Let plug

Wa=10-3/100-0

Wa=0.97

Computation for the mass fraction for Wg using this formula

Wg=Co-Ca/Cg-Ca

Let plug in the formula

Wg=3-0/100-0

Wg=0.03

Second step is to convert the mass fraction to Volume Fraction using this formula

Volume Fraction =[Wg/Pg÷(Wa/Pa)+(Wg/Pg)]*100

Let plug in the formula

Volume =[0.03/2.3 ÷(0.97/7.9)+(0.03/2.3)]*100

Volume=[0.0130435÷0.1227848+0.0130435]*100

Volume=[0.0130435÷0.135828]*100

Volume=0.096*100

Volume=9.60%

Therefore the weight fraction of graphite will be 9.60%


Related Questions

An astronomer of 65 kg of mass hikes from the beach to the observatory atop the mountain in Mauna Kea, Hawaii (altitude of 4205 m). By how much (in newtons) does her weight change when she goes from sea level to the observatory?

Answers

Answer:

[tex]0.845\ \text{N}[/tex]

Explanation:

g = Acceleration due to gravity at sea level = [tex]9.81\ \text{m/s}^2[/tex]

R = Radius of Earth = 6371000 m

h = Altitude of observatory = 4205 m

Change in acceleration due to gravity due to change in altitude is given by

[tex]g_h=g(1+\dfrac{h}{R})^{-2}\\\Rightarrow g_h=9.81\times(1+\dfrac{4205}{6371000})^{-2}\\\Rightarrow g_h=9.797\ \text{m/s}^2[/tex]

Weight at sea level

[tex]W=mg\\\Rightarrow W=65\times 9.81\\\Rightarrow W=637.65\ \text{N}[/tex]

Weight at the given height

[tex]W_h=mg_h\\\Rightarrow W_h=65\times 9.797\\\Rightarrow W_h=636.805\ \text{N}[/tex]

Change in weight [tex]W_h-W=636.805-637.65=-0.845\ \text{N}[/tex]

Her weight reduces by [tex]0.845\ \text{N}[/tex].

Determine the size of memory needed for CD recording of a piece of music, which lasts for 26 minutes, is done with a 20-bit Analog-to-Digital Converter (ADC) in stereo (2 channels), at the rate of 44.1 kSa/s, with the compression factor 6 (allow 10% error margin).

Answers

Answer: the size of memory needed for the CD recording is 28.7 MB

Explanation:

so in the case of stereo, the bitrate is;

⇒ 26 × 60 × 44.1 × 10³ × 2

=   137592 × 10³  

for 10 bit

⇒ 137592 × 10³ × 10

= 1375920 × 10³ bits

now divide by 8 (convert to bytes)

⇒ (1375920 × 10³) / 8

= 171,990,000 BYTE

divide by 1000 (convert to kilobytes)

= 171,990,000 / 1000

= 171,990 KILOBYTES

now Given that, the compression ratio is 6      

so  

171,990 / 6

= 28665 KB

we know that. 1 MB = 1000 KB

x MB = 28665 KB

x MB = 28665 / 1000

⇒ 28.665 MB ≈ 28.7 MB

Therefore the size of memory needed for the CD recording is 28.7 MB

How many flip-flop values are complemented in an 8-bit binary ripple counter to reach the next count value after: 0110111 and 01010110?

Answers

Answer:

- Four (4) flip-flop values will complemented

- one (1) flip-flop value will complemented

Explanation:

To find how many flip flop number of bits complemented, we just need to figure out what the next count in the sequence is and find how many bits have changed.

taking a look at the a) 00110111

we need to just 1 to the value,

so

00110111 +  0000001  = 00111000        

So here, only the first four bits are complemented.

Therefore Four (4) flip-flop values will complemented

Next

b) 01010110

we also add 1 to the value

01010110  + 00000001  = 01010111

only the first bit is complemented.

Therefore one (1) flip-flop value will complemented

Which of the following terms describes the path from an electrical source to a switch or plug?
transmitter
circuit breaker
raceway
breaker panel

Answers

Answer:

transmitter hope thus helped!

Explanation:

Raceway is the answer

"A raceway is an enclosed conduit that forms a physical pathway for electrical wiring."

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