The pressure difference across the sensor housing will be "95 kPa".
According to the question, the values are:
Altitude,
Speed,
Pressure,
The temperature will be:
→ ![T = 15.04-[0.00649(9874)]](https://tex.z-dn.net/?f=T%20%3D%2015.04-%5B0.00649%289874%29%5D)
→ 
→ 
now,
→ ![P_o = 101.29[\frac{(-49.042+273.1)}{288.08} ]^{(5.256)}](https://tex.z-dn.net/?f=P_o%20%3D%20101.29%5B%5Cfrac%7B%28-49.042%2B273.1%29%7D%7B288.08%7D%20%5D%5E%7B%285.256%29%7D)
→
hence,
→ The pressure differential will be:
= 
= 
Thus the above solution is correct.
Learn more about pressure difference here:
brainly.com/question/15732832
Since the armature is wave wound, the magnetic flux per pole is 0.0274 Weber.
<u>Given the following data:</u>
- Number of armature conductors = 144 slots
- Number of poles = 4 poles
- Number of parallel paths = 2
To find the magnetic flux per pole:
Mathematically, the emf generated by a DC generator is given by the formula;
× 
<u>Where:</u>
- E is the electromotive force in the DC generator.
- Z is the total number of armature conductors.
- N is the speed or armature rotation in r.p.m.
- P is the number of poles.
- A is the number of parallel paths in armature.
First of all, we would determine the total number of armature conductors:
×
× 
Z = 864
Substituting the given parameters into the formula, we have;
× 
× 
<em>Magnetic flux </em><em>=</em><em> 0.0274 Weber.</em>
Therefore, the magnetic flux per pole is 0.0274 Weber.
Read more: brainly.com/question/15449812?referrer=searchResults
A) chilled water from evaporator
Answer:All of the above
Explanation:
9.62 psi means 497.49 mm of Hg pressure
for (a)19.58 inches is equals to 497.49 mm of Hg
(b)atmospheric pressure is 14.69 psi
vaccum gauge is 9.62psi
absolute pressure is=14.69-9.62=5.07
(c)vaccum means air is sucked and there is negative pressure so it tells about below atmospheric pressure.
thus all are correct
Answer:
The speed of shaft is 1891.62 RPM.
Explanation:
given that
Amplitude A= 0.15 mm
Acceleration = 0.6 g
So
we can say that acceleration= 0.6 x 9.81

We know that

So now by putting the values



We know that
ω= 2πN/60
198.0=2πN/60
N=1891.62 RPM
So the speed of shaft is 1891.62 RPM.