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Vsevolod [243]
2 years ago
15

How to update android 4.4.2 to 5.1 if there isnt any update available​

Engineering
2 answers:
gulaghasi [49]2 years ago
5 0

Answer:

You cant update it if there isin´t any update available lol

Explanation:

dolphi86 [110]2 years ago
4 0
Well simply you can’t the problem is likely to be that the phone is to old !
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The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to th
Ilia_Sergeevich [38]

Answer:

Flow velocity

50.48m/s

Pressure change at probe tip

1236.06Pa

Explanation:

Question is incomplete

The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to the flow. If the differential height between the water columns connected to the two outlets of the probe is 0.126m, determine (a) the flow velocity and (b) the pressure rise at the tip of the probe. The air temperature and pressure in the duct are 352k and 98 kPa, respectively

solution

In this question, we are asked to calculate the flow velocity and the pressure rise at the tip of probe

please check attachment for complete solution and step by step explanation

8 0
3 years ago
What’s the answer???
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3 years ago
What is an air mass?​
kotegsom [21]

Answer:

An air mass is a body of air with horizontally uniform temperature, humidity, and pressure.

Explanation:

Because it is

8 0
3 years ago
Read 2 more answers
A vacuum pump is used to drain a basement of 20 °C water (with a density of 998 kg/m3 ). The vapor pressure of water at this tem
lord [1]

Answer:

The maximum theoretical height that the pump can be placed above liquid level is \Delta h=9.975\,m

Explanation:

To pump the water, we need to avoid cavitation. Cavitation is a phenomenon in which liquid experiences a phase transition into the vapour phase because pressure drops below the liquid's vapour pressure at that temperature.  As a liquid is pumped upwards, it's pressure drops. to see why, let's look at Bernoulli's equation:

\frac{\Delta P}{\rho}+g\, \Delta h +\frac{1}{2}  \Delta v^2 =0

(\rho stands here for density, h for height)

Now, we are assuming that there aren't friction losses here. If we assume further that the fluid is pumped out at a very small rate, the velocity term would be negligible, and we get:

\frac{\Delta P}{\rho}+g\, \Delta h  =0

\Delta P= -g\, \rho\, \Delta h

This means that pressure drop is proportional to the suction lift's height.

We want the pressure drop to be small enough for the fluid's pressure to be always above vapour pressure, in the extreme the fluid's pressure will be almost equal to vapour pressure.

That means:

\Delta P = 2.34\,kPa- 100 \,kPa = -97.66 \, kPa\\

We insert that into our last equation and get:

\frac{ \Delta P}{ -g\, \rho\,}= \Delta h\\\Delta h=\frac{97.66 \, kPa}{998 kg/m^3 \, \, 9.81 m/s^2} \\\Delta h=9.975\,m

And that is the absolute highest height that the pump could bear. This, assuming that there isn't friction on the suction pipe's walls, in reality the height might be much less, depending on the system's pipes and pump.

8 0
3 years ago
Technician A says that if fuel pump pressure is correct, fuel pump volume will be correct as well. Technician B says that a fuel
guajiro [1.7K]

Answer:

Technician B only

Explanation:

hope this helps :)

5 0
2 years ago
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