Capitated contract, engineering homework help

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Home Work 1 1. What is the weight (lb) of 1 cubic foot of water? Solution. 1 gallon of water = 8.3453 lb 1 cubic foot water = 7.48052 gallons 1 cubic foot water = 8.3453x7.48052 = 62.43lb 2. What is the density (slugs/ft3) of water? Solution. Density of water in slugs/ft3 = 1.940 slugs/ ft3 Density of water is 1g/cm3 1g/cm3 = 1.940slugs/ft3 Therefore density of water is 1.940 slugs/ft3 3. A can measuring 4 inches in diameter and 6 inches high is filled with a liquid. If the net weight is 2 pounds, what is the specific weight of the liquid? Solution d= 4 in, h= 6 in, net wt= 2 pounds specific weight = wt of liquid/volume =2lb/(π/4x42x6) = 0.0265 lb/in3 4. What is the specific gravity of kerosene, which has a specific weight of 7.85*103N/m3? Solution. Specific gravity of kerosene = density of kerosene/ density of water Density of water = 9.81kN/m3 Density of kerosene = 7.85kN/m3 = 7.85/9.81 = 0.8002 5. Two capillary tubes are placed vertically in an open container of water. One tube has a diameter of 2mm, and the other a diameter of 1ich. In which tube wills the water rise higher? Solution If z quantity of water is added into the tubes, the tube with a larger diameter occupies more space than the tube with a lower diameter. Therefore a tube with a lesser diameter will have a higher height than a tube with a larger diameter. Hence d2 has more height than d1 6. A liquid has an absolute viscosity of 2.2*10-5 lb-s/ft2. It weight 45 lb/ft3 what is its kinematic viscosity? Solution Kinematic viscosity v= µ/p Where µ= absolute viscosity P= density Absolute viscocity = 2.2x10-5 lb-s/ft2 Wt= 45 lb/ft3 V = 2.2x10-5/45 =4.89X10-7ft2/lbs 7. A liquid has an absolute viscosity of 2.4*10-5 N-s/m2. It weight 7850 N/m3 what is its kinematic viscosity? Solution. V= 2.4x10-5/7850 = 3.05x10-9 m2/s 8. A plate measuring 20 cm × 20 cm is pulled horizontally through SAE 10 oil at V =0.15 m/s as shown in Figure. The oil temperature is 40oC. Find the force F. The dynamic viscosity of SEA 10 oil is μ = 34 × 10-3 N.s/m2 Solution. F1= s.f at upper side of thin plate F2 = s.f at lower side of thin plate Area of the plate = 0.2x0.2 = 0.04m2 µ= 34x10-3 Ns/m2 therefore newton’s low of viscosity T= µ(ϐu/ϐy) T1 = 34x10-3x (0.15/0.002) T1=2.55N/m2 F1 = T1xA =2.55x0.04 F1 = 0.102N F1= F2 = 0.102 F = 0.102 + 0.102 F = 0.204N 9. The plate in Problem 8 is moved with a velocity of 0.15 m/s to the right but the top and bottom plates are moved to the left at 0.10 m/s, as shown in Figure. Find the force F. Solution. T1 = µ(ϐu/ϐy) T1 = 34x10-3x [(0.15-0.10)/0.002] T1= 0.85N F1 = T1xA =0.85x(0.2x0.2) F1 = 0.034N F1=F2 =0.034N F = F1+F2 =0.068N 10 A capillary tube, having an inside diameter of 6 mm is dipped in water at 20°C. Determine the height of water, which will rise in the tube. Take specific weight of water at 20°C = 998 kg/m3. Take surface tension (σ) = 0.08 gm/cm and angle of contact (α) as 60°. Solution. Capillary rise= (4x0.08x10-3x102 )/(998x9.7x6x10-3) =0.534mm
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Q1. Capitated contract is a healthcare plan that allows the payment of a flat fee for each patient
covered. A managed care organization pays a fixed amount for its members to the providers of
emergency medical services regardless of the number of times the patient is offered the services
(Sandahl et al, 2013).The service provider agreement with the HMO allows the sharing of
deficits and the surplus. Efficient utilization of emergency services may result in a surplus that
the service provider will share with the health plan.
Fixed cost is a cost that doesn’t increase or decrease with any change in amount of services
offered. The expenses have to be paid by an organization independent of any business activity. A
decrease in the volume of services offered produces disproportionately higher decrease in profits
while an increase in volume of services provides the emergency medical service providers with
more profits after the organization achieves the breakeven point.

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Line item budgeting involves grouping of individual financial statements by departments or by
cost centers. It offers the EMS providers a step-by-step approach to budgeting. The providers can
make specific decisions like changing funding levels of various programs to provide money to
more important programs (Schall et al, 2012). The main issue with the budgeting method is the
temptation to rush into unnecessar...


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