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steposvetlana [31]
3 years ago
9

Any smart engineer wanna help me on my test tommorow. Im a sophomore in highschool. Its on logic, and truth tables

Engineering
1 answer:
mojhsa [17]3 years ago
4 0

Answer:

ye sure why not im a jounior about to be a seniour so hit me with the question

Explanation:

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If you owned a business, what are some of the ways you could follow green computing recommendations with regards to recycling an
krek1111 [17]

Answer: E-cycle used computer equipment. Recycle Printer Cartridges and Drum Units. Properly Dispose of Old Technology.

Hope this helps!

8 0
4 years ago
Underground water is to be pumped by a 78% efficient 5- kW submerged pump to a pool whose free surface is 30 m above the undergr
maksim [4K]

Answer:

a) The maximum flowrate of the pump is approximately 13,305.22 cm³/s

b) The pressure difference across the pump is approximately 293.118 kPa

Explanation:

The efficiency of the pump = 78%

The power of the pump = 5 -kW

The height of the pool above the underground water, h = 30 m

The diameter of the pipe on the intake side = 7 cm

The diameter of the pipe on the discharge side = 5 cm

a) The maximum flowrate of the pump is given as follows;

P = \dfrac{Q \cdot \rho \cdot g\cdot h}{\eta_t}

Where;

P = The power of the pump

Q = The flowrate of the pump

ρ = The density of the fluid = 997 kg/m³

h = The head of the pump = 30 m

g = The acceleration due to gravity ≈ 9.8 m/s²

\eta_t = The efficiency of the pump = 78%

\therefore Q_{max} = \dfrac{P \cdot \eta_t}{\rho \cdot g\cdot h}

Q_{max} = 5,000 × 0.78/(997 × 9.8 × 30) ≈ 0.0133 m³/s

The maximum flowrate of the pump Q_{max} ≈ 0.013305 m³/s = 13,305.22 cm³/s

b) The pressure difference across the pump, ΔP = ρ·g·h

∴ ΔP = 997 kg/m³ × 9.8 m/s² × 30 m = 293.118 kPa

The pressure difference across the pump, ΔP ≈ 293.118 kPa

6 0
3 years ago
Carbon dioxide at a temperature of 0oC and a pressure of 600 kPa (abs) flows through a horizontal 40-mm- diameter pipe with an a
rusak2 [61]

Answer:

f = 0.04042

Explanation:

temperature = 0°C = 273k

p = 600 Kpa

d = 40 millemeter

e = 10 m

change in  P = 235 N/m²

μ = 2m/s

R = 188.9 Nm/kgk

we solve this using this formula;

P = ρcos*R*T

we put in the values into this equation

600x10³ = ρcos * 188.9 * 273

600000 = ρcos51569.7

ρcos = 600000/51569.7

=11.63

from here we find the head loss due to friction

Δp/pg = feμ²/2D

235/11.63 = f*10*4/2*40x10⁻³

20.21 = 40f/0.08

20.21*0.08 = 40f

1.6168 = 40f

divide through by 40

f = 0.04042

5 0
3 years ago
Which of the following is not one of the systems required to ensure the safe and correct operation of an engine?
velikii [3]
Brake system

Explanation: the engine doesn’t need to be running to make the brake system work the brake system it’s independent
6 0
3 years ago
HELP ME PLEASE RN
IRISSAK [1]

Answer:

information

Explanation:

see picture

8 0
3 years ago
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