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3 years ago

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? *

VARVARA [1.3K]

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.

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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

Read 2 more answers

There are four spheres of earth. These include __________, _________, ________ and _________. They are important because when th

ololo11 [35]

Answer:

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

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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