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andreyandreev [35.5K]
3 years ago
11

Clean dry and wet mops monthly. O True O False

Engineering
2 answers:
NeTakaya3 years ago
3 0

Answer:

False.

Explanation:

I believe it is false. I hope this helps.

MissTica3 years ago
3 0
True ig (I always clean mine monthly in bleach water)
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Which of the following is the BEST definition for e-commerce? *
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Answer:

(C) Buying and selling items electronically on the internet.

Explanation:

Electronic commerce or e-commerce (sometimes written as eCommerce) is a business model that lets firms and individuals buy and sell things over the internet. E-commerce operates in all four of the following major market segments: ... Business to consumer. Consumer to consumer. Consumer to business.

7 0
3 years ago
Does any1 know TERMS OF MEASUREMENT
brilliants [131]

Answer:

the kilogram (kg), for mass.

the second (s), for time.

the kelvin (K), for temperature.

the ampere (A), for electric current.

the mole (mol), for the amount of a substance.

the candela (cd), for luminous intensity.

the meter (m), for distance.

Explanation:hoped this helped

6 0
3 years ago
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There are four spheres of earth. These include __________, _________, ________ and _________. They are important because when th
ololo11 [35]

Answer:

"lithosphere" , "hydrosphere" , "biosphere" , "atmosphere"

4 0
3 years ago
At what distance from the Earth’s surface is a 10,000 kg satellite if its potential energy is equal to –5.58 x 1011 J? (choose t
Maksim231197 [3]

Answer:

Option D

Explanation:

As we know

Gravitational Constant (G) = 6.67*10^{11}

Mass of Earth (M) = 5.98*10^{24}

Weight of satellite (m) = 10,000 Kg

Radius of Earth (R) = 6.378*10^6

Distance of satellite from Earth = \frac{GMm}{R} - U

Substituting the given values, we get

D=\frac{(-((6.67*10^{11})(5.98*10^{24})(10000))}{(6.378*10^6)} - (-5.58*10^{11})\\\\0.625*10^{3}Km

or 635 m

The correct answer is option D

3 0
3 years ago
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 flo
IRISSAK [1]

The friction loss in the system is 3.480 kilowatts.

<h2>Procedure - Friction loss through a pump</h2><h2 /><h3>Pump model</h3><h3 />

Let suppose that the pump within a distribution system is an open system at steady state, whose mass and energy balances are shown below:

<h3>Mass balance</h3>

\dot m_{in}-\dot m_{out} = 0 (1)

\dot m_{in} = \frac{\dot V_{in}}{\nu_{in}} (2)

\dot m_{out} = \frac{\dot V_{out}}{\nu_{out}} (3)

<h3>Energy balance</h3>

\eta \cdot \dot W_{el} + \dot m_{in}\cdot (h_{in}-h_{out}) - \dot W_{f} = 0 (4)

Where:

  • \dot m_{in} - Inlet mass flow, in kilograms per second.
  • \dot m_{out} - Outlet mass flow, in kilograms per second.
  • \dot V_{in} - Inlet volume flow, in cubic meters per second.
  • \dot V_{out} - Outlet volume flow, in cubic meters per second.
  • \nu_{in} - Inlet specific volume, in cubic meters per kilogram.
  • \nu_{out} - Outlet specific volume, in cubic meters per kilogram.
  • \eta - Pump efficiency, no unit.
  • \dot W_{el} - Electric motor power, in kilowatts.
  • h_{in} - Inlet specific enthalpy, in kilojoules per kilogram.
  • h_{out} - Outlet specific enthalpy, in kilojoules per kilogram.
  • \dot W - Work losses due to friction, in kilowatts.

<h3>Data from steam tables</h3>

From steam tables we get the following water properties at inlet and outlet:

Inlet

p = 100\,kPa, T = 25\,^{\circ}C, \nu = 0.001003\,\frac{kJ}{kg}, h = 104.927\,\frac{kJ}{kg}, Subcooled liquid

Outlet

p = 300\,kPa, T = 25\,^{\circ}C, \nu = 0.001003\,\frac{kJ}{kg}, h = 105.128\,\frac{kJ}{kg}, Subcooled liquid

<h3>Calculation of the friction loss in the system</h3>

If we know that \dot V_{in} = 0.05\,\frac{m^{3}}{s}, \nu_{in} = 0.001003\,\frac{m^{3}}{kg}, h_{in} = 104.927\,\frac{kJ}{kg}, h_{out} = 105.128\,\frac{kJ}{kg}, \eta = 0.90 and \dot W_{el} = 15\,kW, then the friction loss in the system is:

\dot W_{f} = \frac{\dot V_{in}}{\nu_{in}}\cdot (h_{in} - h_{out}) + \eta \cdot \dot W_{el}

\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)

\dot W_{f} = 3.480\,kW

The friction loss in the system is 3.480 kilowatts. \blacksquare

To learn more on pumps, we kindly invite to check this verified question: brainly.com/question/544887

6 0
2 years ago
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