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QveST [7]
3 years ago
6

Does anybody want 20 points? They're free get 'em while ya can...

Engineering
2 answers:
DedPeter [7]3 years ago
4 0

Answer:

hi

Explanation:

Zanzabum3 years ago
4 0

Answer:

i want the points

Explanation:

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It is critical to communicate well during negotiations and conversation in order to attain your objectives. Communication also becomes essential in business and corporation . Effective communication may help you as well as your employees develop a strong professional relationship, which can boost morale and efficiency.

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4 years ago
Consider a N-channel enhancement MOSFET with VGS = 3V, Vt = 1 V, VDS = 10 V, and lambda =0 (channel length modulation parameter)
AveGali [126]

The current IDS is greater than 0 since the VGS has induced an inversion layer and the transistor is operating in the saturation region.

<u>Explanation:</u>

  • Since V_{ds} > V_{gs} - Vt because V_{gs} > Vt.
  • By the saturation region the MOSFET is operating.
  • A specific source voltage and gate of NMOS, the voltage get drained during the specific level, the drain voltage is rises beyond where there is no effect of current during saturated region.
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7 0
3 years ago
3. A multilane highway, with three lanes in each direction, operates at maximum level of services D conditions during the peak h
horsena [70]

Solution :

Given :

The number of lanes in each of the direction is N = 3

The terrain is rolling. And the traffic stream is cars and heavy vehicles. And the peak hour factor is given as 0.85

Now the observed flow speed is S  = 48 mph

Flow rate of vehicles is  V = 3690 veh/h

Highway operates at the maximum LOS (level of service) D condition.

The density value that is allowed at this condition is D = 35 pc/mi/lane

Therefore, $D=\frac{v_p}{S}$

or $v_p = D \times S$

        = 35 x  48

        = 1680 pcphpl

$v_s= \frac{V}{PHF \times N \times f_{HV} \times f_p}$

$1680= \frac{3690}{0.85 \times 3 \times f_{HV} \times 1}$

$f_{HV}=0.8613$

$f_{HV}=\frac{1}{1+P_T(E_T-1)+P_R(E_R-1)}$

$0.8613=\frac{1}{1+P_T(2.5-1)}$

$P_T=0.1073$

So the percentage of trucks is 23.33% of the total volume and is equal to 754 vehicles per hour.

If the heavy vehicles are banned from the highway, the number of vehicles will reduce to 2936 vehicles per hour.

The value of $f_{HV} = 1$

$v_p=\frac{2936}{0.85 \times 3 \times 1 \times 1}$

   = 1151.3 pcphpl

Density value is , $D=\frac{v_p}{S}$

                               = $\frac{1151.3}{48}$

                               23.98 pc/mi/lane

Therefore, the level of service improves to LOS C, when the heavy vehicles are banned in the highways while the car traffic remains the same.

5 0
3 years ago
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