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3 years ago

In highways the far left lane is usually the _____

Engineering

2 answers:

Ivan3 years ago

8 0

Fastest

(Known as the fast lane)

Mice21 [21]3 years ago

8 0

D fastest on most highways

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This elementary problem begins to explore propagation delayand transmission delay, two central concepts in data networking. Cons

telo118 [61]

Explanation:

(a)

Here, distance between hosts A and B is m meters and, propagation speed along the link is s meter/sec

Hence, propagation delay, $d_{prop} = m/sec (s)$

(b)

Here, size of the packet is L bits

And the transmission rate of the link is R bps

Hence, the transmission time of the packet, $d_{trans} = L/R$

(c)

As we know, end-to-end delay or total no delay,

$\mathrm{d}_{\text {nodal }}=\mathrm{d}_{\text {proc }}+\mathrm{d}_{\text {quar }}+d_{\max }+d_{\text {prop }}$

$Here, $\mathrm{d}_{\text {rroc }}$ and $\mathrm{d}_{\text {quat }}$ \\Hence, $\mathrm{d}_{\text {rodal }}=\mathrm{d}_{\text {trass }}+\mathrm{d}_{\text {prop }}$ \\We know, $\mathrm{d}_{\text {trax }}=\mathrm{L} / \mathrm{R}$ sec and $\mathrm{d}_{\text {vapp }}=\mathrm{m} / \mathrm{s}$ sec$ $\text { Hence, } {d_{\text {nodal }}}=\mathrm{L} / \mathrm{R}+\mathrm{m} / \mathrm{s} \text { seconds }$

(d)

The expression, time $time $t=d_{\text {trans }}$$ means the\at time since transmission started is equal to transmission delay.

As we know, transmission delay is the time taken by host to push out the packet.

Hence, at $time $t=d_{\text {trans }}$$ the last bit of the packet has been pushed out or transmitted.

(e)

If $\ d_{prop} >d_{trans}$

Then, at $time $t=d_{\text {trans }}$$ the bit has been transmitted from host A, but to condition (1), the first bit has not reached B.

(f)

If $\ d_{prop}$

Then, at $time $t=d_{\text {trans }}$$ , the first bit has reached destination on B

Here, $s=2.5 \times 10^{8} \mathrm{sec}$

$\begin{aligned}&\mathrm{L}=100 \mathrm{Bits} \text { and }\\&\mathrm{R}=28 \mathrm{kbps} \text { or } 28 \times 1000 \mathrm{bps}\end{aligned}$

It's given that $\ d_{prop} =d_{trans}$

Hence,

$\begin{aligned}\ & \frac{L}{R}=\frac{m}{s} \\m &=s \frac{L}{R} \\&=\frac{2.5 \times 10^{8} \times 100}{28 \times 1000} \\&=892.9 \mathrm{km}\end{aligned}$

5 0

3 years ago

An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no

Stells [14]

Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

Analysis:

a. This is reversible adiabatic(i.e isentropic) process and thus $s_{1} =s_{2}$

From the refrigerant table A11-A13

$P_{1} =0.8MPa$ $\left \{ {{ {{v_{1}=v_{g} @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g} @0.8MPa =246.82 kJ/kg } - also {{s_{1}=s_{g} @0.8MPa =0.91853 kJ/kgK } } \right.$

sat vapor

m= $\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339} = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) = 38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}$

$T_{2} =T_{sat @ 0.2MPa} = -10.09^{o} C$

b.) We take the content of the cylinder as the sysytem.

This is a closed system since no mass leaves or enters.

Hence, the energy balance for adiabatic closed system can be expressed as:

$E_{in} - E_{out} =$ ΔE

$w_{b, out} =$ ΔU

$w_{b, out} =m([tex]u_{1} -u_{2$ )

$w_{b, out} =$ workdone during the isentropic process

=5.8491(246.82-219.9)

=5.8491(26.91)

=157.3993

=157.4kJ

4 0

3 years ago

Wind blows on the side of a fully enclosed hospital located on open flat terrain where V= 120 mi/h. Determine the external press

ss7ja [257]

Answer:

The external pressure is p = -21.9 psf or p = -8.85 psf

Explanation:

Given :

Velocity of wind, v = 120 mi / hr

$k_d = k_c =1$ (wind direction factor)

$k _{zt} = 1$ = topographical factor (for flat terrain)

$q_n$ = velocity pressure at height h

$\therefore q_n = 0.00256 k_z k_{zt} k_d v^2$

$q_n = 0.00256 \times k_z (1)(1)(120)^2$

$= 36.86 k_z$

But for height h = 30 ft, $k_z$ = 0.98 (from table)

$\therefore q_n = 36.86 \times 0.98$

= 36.16

Now, $\frac{L}{B}= \frac{200}{200} =1$ , so $C_p=-0.5$ (from table)

$p = q(G)(C_p)-q_n(GC_{pi})$

where, p = external pressure

G = 0.85 = gust factor (for typical rigid building)

$GC_{pi} = \pm 0.18$ (internal pressure co efficient)

Therefore putting the values,

$p = (36.13)(0.85)(-0.5)-(36.13)(\pm 0.18)$

p = -21.9 psf or p = -8.85 psf

4 0

4 years ago

Why is it important to review your plan when you write an algorithm?

user100 [1]

An algorithm is itself a general step-by-step solution of your problem. ... The most important point here is that you must use algorithms to solve problem, one way or the other. Most of the time it's better to think about your problem before you jump to coding - this phase is often called design.

7 0

3 years ago

1) Find the time in seconds to reach full charge in an RL circuit with L = 5 H and R = 100 ohms

pav-90 [236]

The time constant to reach full charge in an RL circuit is 0.05 ms.

Explanation:

To find the time constant,

The time constant for an RL circuit is defined by τ = L/R.

The given data is

L= 5 H

R= 100 ohms

by using the formula,

τ = L/R

= 5/100

= 0.05 ms

τ = 0.05 ms

Thus, the time constant to reach full charge in an RL circuit is 0.05 ms.

8 0

3 years ago