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.

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].

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

Answer:

transmitter hope thus helped!

Explanation:

Raceway is the answer

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

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