While, Do...while, and For loops can best be characterized as which of the following?

3. (a) (5 points) Suppose N packets arrive simultaneously to a link at which no packets are currently being transmitted or queue

Bezzdna [24]

Answer:

(N-1) × (L/2R) = (N-1)/2

Explanation:

let L is length of packet

R is rate

N is number of packets

then

first packet arrived with 0 delay

Second packet arrived at = L/R

Third packet arrived at = 2L/R

Nth packet arrived at = (n-1)L/R

Total queuing delay = L/R + 2L/R + ... + (n - 1)L/R = L(n - 1)/2R

Now

L / R = (1000) / (10^6 ) s = 1 ms

L/2R = 0.5 ms

average queuing delay for N packets = (N-1) * (L/2R) = (N-1)/2

the average queuing delay of a packet = 0 ( put N=1)

4 0

3 years ago

The alternator produces direct current, which is changed to alternating current by the diodes.

exis [7]

Answer:

False

Explanation:

<u>Alternat</u>or produces <u>alternat</u>ing current ....diodes convert this to direct current

8 0

2 years ago

Which of the following statements are true concerning AC circuit that contains both resistance and inductance? A. The current an

maw [93]

The current will lag the voltage in AC circuit that contains both resistance and inductance.

Answer: C

Explanation

There is no inductance only circuits in reality.

The circuits containing inductance has also a lower amount of resistance.

The current flows in both resistance and inductance.

There is a drop in the total voltage in resistance and inductance giving rise to the voltage applied in the coil when connected in a series.

An example being inductance coil an AC circuit connected to both resistance and inductance in series.

From the vector diagram, this conclusion can be drawn.

3 0

3 years ago

Copper spheres of 20-mm diameter are quenched by being dropped into a tank of water that is maintained at 280 K . The spheres ma

Ivenika [448]

Answer:

The height of the water is 1.25 m

Explanation:

copper properties are:

Kc=385 W/mK

D=20x10^-3 m

gc=8960 kg/m^3

Cp=385 J/kg*K

R=10x10^-3 m

Water properties at 280 K

pw=1000 kg/m^3

Kw=0.582

v=0.1247x10^-6 m^2/s

The drag force is:

$F_{D} =\frac{1}{2} Co*p_{w} A*V^{2}$

The bouyancy force is:

$F_{B} =V*p_{w} *g$

The weight is:

$W=V*p_{c} *g$

Laminar flow:

$v_{T} =\frac{p_{c}-p_{w}*g*D^{2} }{18*u} =\frac{(8960-1000)*9.8*(20x10^{-3})^{2} }{18*0.00143} =1213.48 m/s$

Reynold number:

$Re=\frac{1000*1213.48*20x10^{-3} }{0.00143} \\Re>>1$

Not flow region

For Newton flow region:

$v_{T} =1.75\sqrt{(\frac{p_{c}-p_{w} }{p_{w} })gD }=1.75\sqrt{(\frac{8960-1000}{1000} )*9.8*20x10^{-3} } =2.186m/s$

$Re=\frac{1000*2.186*20x10^{-3} }{0.00143} =30573.4$

$Pr=\frac{\frac{u}{p} }{\frac{K}{pC_{p} } } =\frac{u*C_{p} }{k} =\frac{0.0014394198}{0.582} =10.31$

$Nu=2+(0.4Re^{1/2} +0.06Re^{2/3} )Pr^{2/5} (u/us)^{1/4} \\Nu=2+(0.4*30573.4^{1/2}+0.06*30573.4^{2/3} )*10.31^{2/5} *(0.00143/0.00032)^{1/4} \\Nu=476.99$

$Nu=\frac{h*d}{K_{w} } \\h=\frac{476.99*0.582}{20x10^{-3} } =13880.44W/m^{2} K$

$\frac{T-T_{c} }{T_{w}-T_{c} } =e^{-t/T} \\T=\frac{m_{c}C_{p} }{hA_{c} } =\frac{8960*10x10^{-3}*385 }{13880.44*3} =0.828 s$

$e^{-t/0.828} =\frac{320-280}{360-280} \\t=0.573\\heightofthewater=2.186*0.573=1.25m$

8 0

3 years ago

Which type of engineering design uses an already existing design

Snezhnost [94]

Answer:

Reverse engineering

Explanation:

Reverse Engineering is the remaking of already made products following the deconstruction and examination of the product to make known the product design, code and architecture features, gain knowledge of the composition and construction in a scientific research approach

Reverse engineering is also known as back engineering and consists of three main stages

1) Recovery implementation

2) Design recovery

3) Recovery analysis.

3 0

3 years ago