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2 years ago

Worth 20 points! Please help ASAP!

Engineering

1 answer:

gulaghasi [49]2 years ago

8 0

Answer:

I am in 6th grade, why are high school things popping up??

Explanation:

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Determine the pressure difference in N/m2,between two points 800m apart in horizontal pipe-line,150 mm diameter, discharging wat

zlopas [31]

Answer: $10.631\times 10^3\ N/m^2$

Explanation:

Given

Discharge is $Q=12.5\ L$

Diameter of pipe $d=150\ mm$

Distance between two ends of pipe $L=800\ m$

friction factor $f=0.008$

Average velocity is given by

$\Rightarrow v_{avg}=\dfrac{12.5\times 10^{-3}}{\frac{\pi }{4}(0.15)^2}\\\\\Rightarrow v_{avg}=\dfrac{15.9134\times 10^{-3}}{2.25\times 10^{-2}}\\\\\Rightarrow v_{avg}=7.07\times 10^{-1}\\\Rightarrow v_{avg}=0.707\ m/s$

Pressure difference is given by

$\Rightarrow \Delta P=f\ \dfrac{L}{d}\dfrac{\rho v_{avg}^2}{2}\\\\\Rightarrow \Delta P=0.008\times \dfrac{800}{0.15}\times \dfrac{997\times (0.707)^2}{2}\\\\\Rightarrow \Delta P=10,631.45\ N/m^2\\\Rightarrow \Delta P=10.631\ kPa$

8 0

2 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

2 years ago

1. A thin plate of a ceramic material with E = 225 GPa is loaded in tension, developing a stress of 450 MPa. Is the specimen lik

mina [271]

Answer:

fracture will occur as the value is less than E/10 (= 22.5)

Explanation:

If the maximum strength at tip Is greater than theoretical fracture strength value then fracture will occur and if the maximum strength is lower than theoretical fracture strength then no fracture will occur.

$\sigma_m = 2\sigma_o [\frac{a}{\rho_t}]^{1/2}$