AP Physics B help please on fluids in motion. could some1 plz help me with these questions?

Question

42. A liquid (p = 1.65 g/cm^3) flows through two horizontal sections of tubing joined end to end. In the first section the
cross-sectional area is 10.0 cm^2, the flow speed is 275 cm/ s. and the pressure is 1.20 * 10^5 Pa. In the second section
the cross-sectional area is 2.50 cm^2. Calculate the smaller section's (a) flow speed and (b) pressure.

43. A hypodermic syringe contains a medicine with the den-SM sity of water. The barrel of the svringe has a cross-sectional
area of 2.50 X 1O^-5 m^2. In the absence of a force on the plunger, the pressure everywhere is 1.00 atm. A force F of
magnitude 2.00 N is exerted on the plunger, making medicine squirt from the needle. Determine the medicine's flow speed
through the needle. Assume that the pressure in the needle remains equal to 1.00 atm and that the syringe is horizontal.

44. When a person inhales, air moves down the bronchus (windpipe) at 15 cm/s. The average flow speed of the air
doubles through a constriction in the bronchus. Assuming incompressible flow, determine the pressure drop in the
constriction.

Answer 1

42.
(a) v2 = (10.0 cm^2)(275 cm/s)/(2.50 cm^2)
v2 = 1,100 cm/s
(b) P2 = P1 - (ρ/2)(v2^2 - v1^2)
P2 = 1.20*10^5 Pa - (825 kg/m^3)(11^2 m^2 - 2.75^2 m^2)
P2 = 1.20*10^5 Pa - (825 kg/m^3)(121 m^2 - 7.5625 m^2)
P2 = 26,414.0625 ≈ 26.4 kPa

43.
v^2 = (2)(2 N)/[(2.50*10^-5 m^2)(10^3 kg/m^3)]
v ≈ 12.649 m/s ≈ 12.6 m/s

44.
∆P = (ρ/2)(v2^2 - v1^2)
∆P = (ρ/2)(4*0.0225 - 0.0225)
∆P = 0.03375ρ