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2 years ago

Sinks must be used for the correct intended purpose to prevent

Engineering

1 answer:

irakobra [83]2 years ago

8 0

Answer:

... spilling water or getting anything cascading onto the floor

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I’m doing a project on renewable energy. There are 6 energy sources. Solar, wind, geothermal, hydroelectric, tidal, and biomass.

nalin [4]

Answer:

"Biofuels"

Explanation:

I don't know if this counts but I guess it's not one of those.

6 0

2 years ago

Read 2 more answers

A closed, rigid tank is lled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. The gas is he

sergejj [24]

Answer:

T₂ =93.77 °C

Explanation:

Initial temperature ,T₁ =27°C= 273 +27 = 300 K

We know that

Absolute pressure = Gauge pressure + Atmospheric pressure

Initial pressure ,P₁ = 300+1=301 kPa

Final pressure ,P₂= 367+1 = 368 kPa

Lets take temperature=T₂

We know that ,If the volume of the gas is constant ,then we can say that

$\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}$

${T_2}=T_1\times \dfrac{P_2}{P_1}$

Now by putting the values in the above equation we get

${T_2}=300\times \dfrac{368}{301}\ K$

The temperature in °C

T₂ = 366.77 - 273 °C

T₂ =93.77 °C

8 0

3 years ago

What does this work for

Anastaziya [24]

Answer:

it allows your dash board to light up you MPH RPM and all the other numbers on the spadomter

Explanat

8 0

2 years ago

A large heat pump should upgrade 5 MW of heat at 85°C to be delivered as heat at 150°C. Suppose the actual heat pump has a COP o

AysviL [449]

Answer:

W=2 MW

Explanation:

Given that

COP= 2.5

Heat extracted from 85°C

Qa= 5 MW

Lets heat supplied at 150°C = Qr

The power input to heat pump = W

From first law of thermodynamics

Qr= Qa+ W

We know that COP of heat pump given as

$COP=\dfrac{Qr}{W}$

$2.5=\dfrac{5}{W}$

W=2 MW

For Carnot heat pump

$COP=\dfrac{T_2}{T_2-T_1}$

$2.5=\dfrac{T_2}{T_2-(273+85)}$

2.5 T₂ - 895= T₂

T₂=596.66 K

T₂=323.6 °C

7 0

2 years ago

2.) A fluid moves in a steady manner between two sections in a flow

Talja [164]

Answer:

$250\ \text{lbm/min}$

$625\ \text{ft/min}$

Explanation:

$A_1$ = Area of section 1 = $10\ \text{ft}^2$

$V_1$ = Velocity of water at section 1 = 100 ft/min

$v_1$ = Specific volume at section 1 = $4\ \text{ft}^3/\text{lbm}$

$\rho$ = Density of fluid = $0.2\ \text{lb/ft}^3$

$A_2$ = Area of section 2 = $2\ \text{ft}^2$

Mass flow rate is given by

$m=\rho A_1V_1=\dfrac{A_1V_1}{v_1}\\\Rightarrow m=\dfrac{10\times 100}{4}\\\Rightarrow m=250\ \text{lbm/min}$

The mass flow rate through the pipe is $250\ \text{lbm/min}$

As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1

$m=\rho A_2V_2\\\Rightarrow V_2=\dfrac{m}{\rho A_2}\\\Rightarrow V_2=\dfrac{250}{0.2\times 2}\\\Rightarrow V_2=625\ \text{ft/min}$

The speed at section 2 is $625\ \text{ft/min}$ .

3 0

2 years ago