11. As __and___ prices continued to rise in the late 1960’s and 70's, 4 and 6 cylinder engines began to make a comeback.

One method that is used to grow nanowires (nanotubes with solid cores) is to initially deposit a small droplet of a liquid catal

7nadin3 [17]

Answer: maximum length of the nanowire is 510 nm

Explanation:

From the table of 'Thermo physical properties of selected nonmetallic solids at At T = 1500 K

Thermal conductivity of silicon carbide k = 30 W/m.K

Diameter of silicon carbide nanowire, D = 15 x 10⁻⁹ m

lets consider the equation for the value of m

m = ( (hP/kAc)^1/2 ) = ( (4h/kD)^1/2 )

m = ( ((4 × 10⁵)/(30×15×10⁻⁹ ))^1/2 ) = 942809.04

now lets find the value of h/mk

h/mk = 10⁵ / ( 942809.04 × 30) = 0.00353

lets consider the value θ/θb by using the equation

θ/θb = (T - T∞) / (T - T∞)

θ/θb = (3000 - 8000) / (2400 - 8000)

= 0.893

the temperature distribution at steady-state is expressed as;

θ/θb = [ cosh m(L - x) + ( h/mk) sinh m (L - x)] / [cosh mL+ (h/mk) sinh mL]

θ/θb = [ cosh m(L - L) + ( h/mk) sinh m (L - L)] / [cosh mL+ (h/mk) sinh mL]

θ/θb = [ 1 ] / [cosh mL+ (h/mk) sinh mL]

so we substitute

0.893 = [ 1 ] / [cosh (942809.04 × L) + (0.00353) sinh (942809.04 × L)]

L = 510 × 10⁻⁹m

L = 510 nm

therefore maximum length of the nanowire is 510 nm

4 0

2 years ago

3. In the text, we described a multithreaded file server, showing why it is better than a single-threaded server and a finite-st

NemiM [27]

Answer and Explanation:

• 1 thread awaits the incoming request

• 1 thread responds to the request

• 1 thread reads the hard disk

A multithreaded file server is better than a single-threaded server and a finite-state machine server because it provides better response compared to the rest and can make use of the shared Web data.

Yes, there are circumstances in which a single-threaded server might be better. If it is designed such that:

- the server is completely CPU bound, such that multiple threads isn't needed. But it would account for some complexity that aren't needed.

An example is, the assistance number of a telephone directory (e.g 7771414) for an community of say, one million people. Consider that each name and telephone number record is sixty-four characters, the whole database takes 64 MB, and can be easily stored in the server's memory in order to provide quick lookup.

NOTE:

Multiple threads lead to operation slow down and no support for Kernel threads.

4 0

3 years ago

A thick steel slab (rho= 7800 kg/m3 , cp= 480 J/kg K, k= 50 W/m K) is initially at 300 °C and is cooled by water jets impinging

dimaraw [331]

Answer:

t = 2244.3 sec

Explanation:

calculate the thermal diffusivity

$\alpha = \frac{k}{\rho c}$

$= \frac{50}{7800\times 480} = 1.34 \times 10^{-5} m^2/s$

Temperature at 28 mm distance after t time = = 50 degree C

we know that

$\frac[ T_{28} - T_s}{T_i -T_s} = erf(\frac{x}{2\sqrt{at}})$

$\frac{ 50 -25}{300-25} = erf [\frac{28\times 10^{-3}}{2\sqrt{1.34\times 10^{-5}\times t}}]$

$0.909 = erf{\frac{3.8245}{\sqrt{t}}}$

from gaussian error function table , similarity variable w calculated as

erf w = 0.909

it is lie between erf w = 0.9008 and erf w = 0.11246 so by interpolation we have

w = 0.08073

$erf 0.08073 = erf[\frac{3.8245}{\sqrt{t}}]$

$0.08073 = \frac{3.8245}{\sqrt{t}}$

solving fot t we get

t = 2244.3 sec

3 0

2 years ago

Write a python program to apply this:

lakkis [162]

Answer:

# Initialize a dictionary with the keys

contestants = {"Darci Lynne":0, "Angelica Hale":0, "Angelina Green":0};

# Repeatedly prompt the user for a contestant name to vote for

while True:

# Prompting user for contestant name

cName = input("Enter contestant name to vote: ");

# Checking for Done

if cName.lower() == "done":

break;

# Checking in dictionary

if cName in contestants.keys():

# Updating vote value

contestants[cName] += 1

# New entry

else:

contestants[cName] = 1

# Printing header

print("\n%-20s %-15s\n" %("Contestant Name", "Votes Casted"))

# Printing results

for contestant in contestants:

print("%-23s %-15d" %(contestant, contestants[contestant]))

8 0

3 years ago

The mean of hours that the average person watches television each day is 4.18 hours with a standard deviation of 1.19 hours. Fin

luda_lava [24]

Answer:

$z = \frac{3-4.18}{1.19}=-0.992$

$z = \frac{5-4.18}{1.19}=0.689$

And we can find this probability with this difference:

$P(-0.992$

And then we can conclude that the probability that someone watches between 3 and 5 hours a day is approximately 0.591 using a normal distribution

Explanation:

For this case we can define the random variable X as "hours that a person watches television". For this case we don't have the distribution for X but we have the following parameters:

$\mu = 4.18,\sigma =1.19$