A red circle and diagonal slash on a sign means that:.

A rigid tank having 25 m3 volume initially contains air having a density of 1.25 kg/m3, then more air is supplied to the tank fr

Hoochie [10]

Answer:

$\Delta m = 102.25\,kg$

Explanation:

The mass inside the rigid tank before the high pressure stream enters is:

$m_{o} = \rho_{air}\cdot V_{tank}$

$m_{o} = (1.25\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})$

$m_{o} = 31.25\,kg$

The final mass inside the rigid tank is:

$m_{f} = \rho \cdot V_{tank}$

$m_{f} = (5.34\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})$

$m_{f}= 133.5\,kg$

The supplied air mass is:

$\Delta m = m_{f}-m_{o}$

$\Delta m = 133.5\,kg-31.25\,kg$

$\Delta m = 102.25\,kg$

4 0

3 years ago

Hello I need some help with this please. Pick a problem in your school or community that you think could be solved with technolo

bezimeni [28]

Answer:

An AI operated automatic garbage collection system

Explanation:

There is always an issue in my neighbourhood with the garbagemen coming on time so having an automatic system will help in the overall efficiency in the task

7 0

3 years ago

Read 2 more answers

Consider a Carnot refrigeration cycle executed in a closed system in the saturated liquid–vapor mixture region using 1.06 kg of

Alexxandr [17]

Answer:

$P_m_i_n = 442KPA$

Explanation:

We are given:

m = 1.06Kg

$T_H = 1.2T_L$

T = 22kj

Therefore we need to find coefficient performance or the cycle

$COP_R = \frac {1}{(T_R/T_l) -1}$

$= \frac {1 }{1.2-1}$

= 5

For the amount of heat absorbed:

$Q_l = COP_R Wm$

= 5 × 22 = 110KJ

For the amount of heat rejected:

$Q_H = Q_L + W_m$

= 110 + 22 = 132KJ

[tex[ q_H = \frac{Q_L}{m} [/tex];

= $= \frac{132}{1.06}$

= 124.5KJ

Using refrigerant table at hfg = 124.5KJ/Kg we have 69.5°c

Convert 69.5°c to K we have 342.5K

To find the minimum temperature:

$T_L = \frac{T_H}{1.2}$ ;

$T_L = \frac{342.5}{1.2}$

= 285.4K

Convert to °C we have 12.4°C

From the refrigerant R -134a table at $T_L$ = 12.4°c we have 442KPa

6 0

3 years ago

Explain why the failure of a garden hose occurred near its end and why the tear occurred along its length. Use numerical values

Nataliya [291]

Answer:

Most hydraulic systems develops pressure surges that may surpass settings valve. by exposing the hose surge to pressure above the maximum operating pressure will shorten the hose life.

Explanation:

Solution

Almost all hydraulic systems creates pressure surges that may exceed relief valve settings. exposing the hose surge to pressure above the maximum operating pressure shortens the hose life.

In systems where pressure peaks are severe, select or pick a hose with higher maximum operating pressure or choose a spiral reinforced hose specifically designed for severe pulsing applications.

Generally, hoses are designed or created to accommodate pressure surges and have operating pressures that is equal to 25% of the hose minimum pressure burst.

7 0

3 years ago

Currently, the lost time of each stage is 4 seconds, and intersection critical v/c ratio is set to be 0.75 to avoid cycle failur

KIM [24]

Find the solution in the attachments

Note: Question was incomplete, so the complete question is added in the attachments.

7 0

3 years ago