Answer:
the pressure reading when connected a pressure gauge is 543.44 kPa
Explanation:
Given data
tank volume (V) = 400 L i.e 0.4 m³
temperature (T) = 25°C i.e. 25°C + 273 = 298 K
air mass (m) = 3 kg
atmospheric pressure = 98 kPa
To find out
pressure reading
Solution
we have find out pressure reading by gauge pressure
i.e. gauge pressure = absolute pressure - atmospheric pressure
first we find absolute pressure (p) by the ideal gas condition
i.e pV = mRT
p = mRT / V
p = ( 3 × 0.287 × 298 ) / 0.4
p = 641.44 kPa
so
gauge pressure = absolute pressure - atmospheric pressure
gauge pressure = 641.44 - 98
gauge pressure = 543.44 kPa
Answer:
Explanation:
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Answer:
D. protruding steel rebars .. #answerwithquality #BAL
Answer:
Different types of equipment are required for proper conditioning of air because every air conditional space faces some geometrical and environmental issues or problems. There are some different types of equipment used for conditioning of air that are air system, water system and air-water system. In many cases the air conditioning of the system varies with size of the equipment.
Answer:
Electroosmotic velocity will be equal to 
Explanation:
We have given applied voltage v = 100 volt
Length of capillary L = 5 mm = 0.005 m
Zeta potential of the capillary surface 
Dielectric constant of glass is between 5 to 10 here we are taking dielectric constant as 
Viscosity of glass is 
Electroosmotic velocity is given as 
So Electroosmotic velocity will be equal to
