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:
→ [tex]T = 15.04-[0.00649(9874)][/tex]
→ [tex]= 15.04-64.082[/tex]
→ [tex]= -49.042^{\circ} C[/tex]
now,
→ [tex]P_o = 101.29[\frac{(-49.042+273.1)}{288.08} ]^{(5.256)}[/tex]
→ [tex]= 27.074[/tex]
hence,
→ The pressure differential will be:
= [tex]122-27[/tex]
= [tex]95 \ kPa[/tex]
Thus the above solution is correct.
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