Answer:
There are many life-saving benefits of biotechnology.
Explanation:
Explain why power is generated at very high voltages.
Answer:
It's because to increase efficiency.
Explanation:
As electricity is transmitted over long distances, there are inherent energy losses along the way.
(I hope this will help)
innovative ideas for civil engineering individual project? I'm running out of time. Need to submit and get approval for this. Please help me and give me a new title to research.
For an egsample- Investigation of replacing Ricehusk instead of Sand in C30, Like wise
Answer:
Top Final year projects for civil engineering students
• Geographic Information System using Q-GIS. ...
• Structural and Foundation Analysis. ...
• Construction Project Management & Building Information Modeling. ...
• Tall Building Design. ...
• Seismic Design using SAP2000 & ETABS.
Explanation:
Hope it's help
determine the values of the forces acting at A and B for the forces system
Answer:
You forgot the image and or example. I mean the answer could be anything right now.
Discuss what is an ore.
Answer:
a naturally occurring solid material from which a metal or valuable mineral can be profitably extracted.
Explanation:
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Answer:
Ore is an accumulation of any mineral mixed with other elements. The mineral content of an ore must be in sufficient concentration to make its extraction commercially and economically viable.
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If you have any query, feel free to ask.
When is the output of a NAND-gate HIGH?
Answer:
In digital electronics, a NAND gate (NOT-AND) is a logic gate which produces an output which is false only if all its inputs are true; thus its output is complement to that of an AND gate. A LOW (0) output results only if all the inputs to the gate are HIGH (1); if any input is LOW (0), a HIGH (1) output results.
1. What are some pendulum wave characteristics?
2. What are waves characteristics?
3. List characteristics that they both share
Answer:
Explanation:
An intake manifold gasket has been replaced due to a vacuum leak. Which of the following steps uses a scan tool to complete the job? O A. Torquing the manifold bolts B. Idle relearn O C. Refilling the cooling system O D. Air cleaner check
Answer: B.Idle relearn
Explanation:
When replacing an intake manifold gasket due to a vacuum leak, using a scan tool for idle relearn is a crucial step to complete the job. The Option B.
How does a scan tool help complete the job when replacing an intake manifold gasket?The scan tool is used to reset the idle control system and allow the engine's computer to relearn the correct idle speed and air/fuel mixture. This is important because the replacement of the intake manifold gasket can affect the engine's idle characteristics.
By using the scan tool to perform an idle relearn procedure, the engine management system can recalibrate and optimize the idle control parameters, ensuring smooth and stable idle operation. This step helps to restore the engine's performance and maintain proper combustion efficiency after the intake manifold gasket replacement.
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When a tensile specimen is stretched in the plastic region to an engineering strain of 0.2, calculate the amount of cold work percent.
Answer:
0.2 x 100
Explanation:
Engineering strain is the original crossection/original crossection
cold work percentage is
original crossection/original crossection x 100
What is The output of full-wave rectifier with filter
Answer:
The output we get from a full-wave rectifier is a pulsating DC voltage that increases to a maximum and then decreases to zero.
Explanation:
hope it helps sorry if im wrong
The product of two factors is 4,500. If one of the factors is 90, which is the other factor?
Compare the output of full-wave rectifier with and without filter
Answer:
Full wave rectification flips the negative half cycle of the sine wave to positive so the result is two positive half cycles.
Explanation:
hope it helps a lil
A pump of a water distribution system at 25°C is powered by a 15 kW electric motor whose efficiency is 90 percent. The water flow rate through the pump is 50 L/s. The diameters of the inlet and outlet pipes are both 5 cm and the elevation difference across the pump is negligible. If the absolute pressures at the inlet and outlet of the pump are measured as 100 kPa and 300 kPa, respectively, determine the friction loss in the system.
The friction loss in the system is 3.480 kilowatts.
Procedure - Friction loss through a pumpPump modelLet suppose that the pump within a distribution system is an open system at steady state, whose mass and energy balances are shown below:
Mass balance[tex]\dot m_{in}-\dot m_{out} = 0[/tex] (1)
[tex]\dot m_{in} = \frac{\dot V_{in}}{\nu_{in}}[/tex] (2)
[tex]\dot m_{out} = \frac{\dot V_{out}}{\nu_{out}}[/tex] (3)
Energy balance
[tex]\eta \cdot \dot W_{el} + \dot m_{in}\cdot (h_{in}-h_{out}) - \dot W_{f} = 0[/tex] (4)
Where:
[tex]\dot m_{in}[/tex] - Inlet mass flow, in kilograms per second.[tex]\dot m_{out}[/tex] - Outlet mass flow, in kilograms per second. [tex]\dot V_{in}[/tex] - Inlet volume flow, in cubic meters per second. [tex]\dot V_{out}[/tex] - Outlet volume flow, in cubic meters per second. [tex]\nu_{in}[/tex] - Inlet specific volume, in cubic meters per kilogram.[tex]\nu_{out}[/tex] - Outlet specific volume, in cubic meters per kilogram.[tex]\eta[/tex] - Pump efficiency, no unit.[tex]\dot W_{el}[/tex] - Electric motor power, in kilowatts.[tex]h_{in}[/tex] - Inlet specific enthalpy, in kilojoules per kilogram.[tex]h_{out}[/tex] - Outlet specific enthalpy, in kilojoules per kilogram. [tex]\dot W[/tex] - Work losses due to friction, in kilowatts. Data from steam tablesFrom steam tables we get the following water properties at inlet and outlet:
Inlet
[tex]p = 100\,kPa[/tex], [tex]T = 25\,^{\circ}C[/tex], [tex]\nu = 0.001003\,\frac{kJ}{kg}[/tex], [tex]h = 104.927\,\frac{kJ}{kg}[/tex], Subcooled liquid
Outlet
[tex]p = 300\,kPa[/tex], [tex]T = 25\,^{\circ}C[/tex], [tex]\nu = 0.001003\,\frac{kJ}{kg}[/tex], [tex]h = 105.128\,\frac{kJ}{kg}[/tex], Subcooled liquid
Calculation of the friction loss in the systemIf we know that [tex]\dot V_{in} = 0.05\,\frac{m^{3}}{s}[/tex], [tex]\nu_{in} = 0.001003\,\frac{m^{3}}{kg}[/tex], [tex]h_{in} = 104.927\,\frac{kJ}{kg}[/tex], [tex]h_{out} = 105.128\,\frac{kJ}{kg}[/tex], [tex]\eta = 0.90[/tex] and [tex]\dot W_{el} = 15\,kW[/tex], then the friction loss in the system is:
[tex]\dot W_{f} = \frac{\dot V_{in}}{\nu_{in}}\cdot (h_{in} - h_{out}) + \eta \cdot \dot W_{el}[/tex]
[tex]\dot W_{f} = \left(\frac{0.05\,\frac{m^{3}}{s} }{0.001003\,\frac{m^{3}}{kg} } \right)\cdot \left(104.927\,\frac{kJ}{kg}-105.128\,\frac{kJ}{kg}\right) + (0.90)\cdot (15\,kW)[/tex]
[tex]\dot W_{f} = 3.480\,kW[/tex]
The friction loss in the system is 3.480 kilowatts. [tex]\blacksquare[/tex]
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