Which of the following technologies might affect how a vehicle is driven

A reservoir delivers water to a horizontal pipeline 39 long The first 15 m has a diameter of 50 mm, after which it suddenly beco

allsm [11]

Answer:

The difference of head in the level of reservoir is 0.23 m.

Explanation:

For pipe 1

$d_1=50 mm,f_1=0.0048$

For pipe 2

$d_2=75 mm,f_2=0.0058$

Q=2.8 l/s

$Q=2.8\times 10^{-3]$

We know that Q=AV

$Q=A_1V_1=A_2V_2$

$A_1=1.95\times 10^{-3}m^2$

$A_2=4.38\times 10^{-3} m^2$

$So V_2=0.63 m/s,V_1=1.43 m/s$

head loss (h)

$h=\dfrac{f_1L_1V_1^2}{2gd_1}+\dfrac{f_2L_2V_2^2}{2gd_2}+0.5\dfrac{V_1^2}{2g}$

Now putting the all values

$h=\dfrac{0.0048\times 15\times 1.43^2}{2\times 9.81\times 0.05}+\dfrac{0.0058\times 24\times 0.63^2}{2\times 9.81\times 0.075}+0.5\dfrac{1.43^2}{2\times 9.81}$

So h=0.23 m

So the difference of head in the level of reservoir is 0.23 m.

8 0

3 years ago

An LED camping headlamp can run for 18 hours, powered by three AAA batteries. The batteries each have a capacity of 1000 mAh, an

KIM [24]

Answer:

a) the power consumption of the LEDs is 0.25 watt

b) the LEDs drew 0.0555 Amp current

Explanation:

Given the data in the question;

Three AAA Batteries;

<---- 1000mAh [ + -] 1.5 v ------1000mAh [ + -] 1.5 v --------1000mAh [ + -] 1.5 v------

so V_total = 3 × 1.5 = 4.5V

a) the power consumption of the LEDs

I_battery = 1000 mAh / 18hrs { for 18 hrs}

I_battery = 1/18 Amp { delivery by battery}

so consumption by led = I × V_total

we substitute

⇒ 1/18 × 4.5

P = 0.25 watt

Therefore the power consumption of the LEDs is 0.25 watt

b) How much current do the LEDs draw

I_Draw = I_battery = 1/18 Amp = 0.0555 Amp

Therefore the LEDs drew 0.0555 Amp current

5 0

3 years ago

What do you think of this schematic diagram?

Sholpan [36]

Explanation:

A schematic diagram is a picture that represents the components of a process, device, or other object using abstract, often standardized symbols and lines. ... Schematic diagrams do not include details that are not necessary for comprehending the information that the diagram was intended to convey.

5 0

3 years ago

A westbound section of freeway currently has three 12-ft wide lanes, a 6-ft right shoulder, and no ramps within 3 miles upstream

Tresset [83]

Answer:

•Estimated density = 39.685Pc/mi/en

•Level of service, LOS frequency = LOSC

Explanation:

We are given:

•Freeway current lane width,B = 12ft

• freeway current shoulder width,b = 6ft

• percentage of heavy vehicle, Ptb = 10℅

• peak hour factor, PHF = 0.9

Let's consider,

•Number of lanes N = 4

• flow of traffic V = 7500vph

• percentage of Rv = 0, therefore the freeflow speed in freeway FFS = 70mph

• cars equivalent for recreational purpose Er= 2

•cars to be used for trucks and busses Etb= 2.5

Let's first calculate for the heavy adjustment factor.

We have:

$F_H_v = \frac{1}{1+P_t_b(E_t_b-1)+Pr(Er-1)}$

Substituting figures in the equation we have:

$= \frac{1}{1+0.1(2.5-1)+0(2-1)}$

= 0.75

Let's now calculate equivalent flow rate of the car using:

$Vp = \frac{V}{(P_H_F)*N)*(F_H_v)*(F_p)}$

$= \frac{7500}{0.9*4*0.75*1}$

= 2777.7 pc/h/en

Calculating for traffic density, we have:

$D = \frac{Vp}{FFS}$

$D = \frac{2777.7}{70}$

D = 39.685 Pc/mi/en

Using the table for LOS criteria of basic frequency segment, the level of service LOS of frequency is LOSC

4 0

3 years ago

The common type of defects found when soldering on a printed circuit board

sammy [17]

Explanation:

Solder Bridges

Plating Voids

Non-wetting or dewetting.

5 0

3 years ago