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Physics Questions:

#1. Water flows steadily from an open tank as shown in the…

Complete Question Text:

#1. Water flows steadily from an open tank as shown in the figure. (Figure 1) The elevation of point 1 is 10.0m , and the elevation of points 2 and 3 is 2.00m . The cross-sectional area at point 2 is 4.80×10−2m2 ; at point 3, where the water is discharged, it is 1.60×10−2m2 . The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe.

A. Assuming that Bernoulli's equation applies, compute the volume of water ΔV that flows across the exit of the pipe in 1.00s . In other words, find the discharge rate ΔV/Δt.

I got .2 m^3/s and that was correct.

B. What is the gauge pressure p_gauge at point 2?

#2. You are on the 48th floor of the Windom Hotel. The day is stormy, and you wonder whether the hotel is a safe place. According to the weather report, the air speed is 22m/s ; the size of the window in your room is 0.50m × 1.9m .

A. What is the difference in pressure between the inside and outside air? Assume the air flow to be laminar (non-turbulent) and non-viscous. The density of air is 1.30 kg/m3.

B. What is the magnitude of the net force that the air exerts on the window?
Express your answer to two significant figures and include the appropriate units.

C. What is the direction of the net force that the air exerts on the window? (INWARD OR OUTWARD?)

#3. Comet crash On June 30, 1908, a monstrous comet fragment of mass greater than 10^9 kg is thought to have devastated a 2000-km2 area of remote Siberia (this impact was called the Tunguska event). Assume the density of the comet is ρcomet = 5500 kg/m3 and the drag coefficient is 0.50.

A. Determine the terminal speed of such a comet in air of density 0.70 kg/m^3.
Express your answer to two significant figures and include the appropriate units.

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Asked on Dec. 10, 2013, 6:22 p.m.
Nishank S. Tutors Physics + 13 others
New Jersey Institute of Technology 2013 - Civil Engineering
48
"I currently tutor College level Statistics and Calculus for Tutor.com. I have been tutoring there for close to two years and..."

Part B:
Use Bernoulli's equation:
z1 + (v1^2)/(2g) + (p1)/(9.81 kN/m^3) = z2 + (v2^2)/(2g) + (p2)/(9.81 kN/m^3)

Since water tank is open at point 1, p1 or the pressure at 1 = 0. Also, the velocity is also 0 at point 1since cross-sectional area of tank is much bigger than the cross-sectional area of the pipe. We will be using our datum at the elevation of points 2 and 3 for Bernoulli's equation so since point 1 is 10 - 2 = 8 m above points 2 and 3, our z1 = 8 m. Also, note that via law of continuity:
Flow at point 2 = Flow at point 3
(Velocity at point 2)(4.8 * 10^-2 m^2) = 0.2 m^3/s
Velocity at point 2 = 4.17 m/s = v2
Note that z2 = 0 since we took our datum about points 2 and 3.
Plugging all of this in and solving for p2:
8 m = (4.17 m/s)^2/(2*9.81 m/s^2) + (p2)/(9.81 kN/m^3)
p2 = 69.78555 KPa = kN/m^2

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