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DENIUS [597]
3 years ago
10

What does pharmacist do in medical science?​

Engineering
1 answer:
Kipish [7]3 years ago
7 0
They are health care professionals that focus on the safe and effective use of medications
You might be interested in
Consider the freeway in Problem 1. At one point along this freeway there is a 4% upgrade with a directional hourly traffic volum
ryzh [129]

Answer:

The Question is incomplete, the complete question is as follows:

<em>Consider the freeway in Problem 1. At one point along this freeway there is a 4% upgrade with a directional hourly traffic volume of 5,435 vehicles. If all other conditions are as described in Problem 1, how long can this grade be without the freeway LOS dropping to F? </em>

A six-lane rural freeway (three lanes in each direction) has regular weekday users and currently operates at maximum LOS C conditions. The base free-flow speed is 65 mi/h, lanes are 11 ft wide, the right-side shoulder is 4 ft wide, and the interchange density is 0.25 per mile. The highway is one rolling terrain with 10% large trucks and buses (no recreational vehicles), and the peak-hour factor is 0.90. Determine the hourly volume for these conditions

Explanation:

<em>Make the assumption Base continuous flow velocity (BFFS)= 65 mph. </em>

Pitch width= 11 ft.

Decrease in lane width pace,fLW= 1.9 mph.

Complete Lateral clearance= 4 ft. Lateral clearance speed reduction, fLC= 0.8 mph.

Complete Width of the Ramp= 0.25 mile.

Velocity reduction proportional to the ramp height, f ID= 0 mph.

Assume lane number to be = 3.

Reduction in speed corresponding to no. of lanes, fN = 3 mph

Free Flow Speed (FFS) = BFFS – fLW – fLC – fN – fID = 65 – 1.9 – 0.8 – 3 – 0 = 59.3 mph

Peak Flow, V veh/hr

Peak-hour factor = 0.90

Trucks = 10%

Rolling Terrain

fHV = 1/ (1 + 0.10 (2.5-1)) = 1/1.15 = 0.8696

fP = 1.0

Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = V/ (0.90*3*0.8696*1.0) = 0.426V veh/hr/ln

Average speed of vehicles, S = FFS = 59.3 mph

Level of service C

Density of LOS C lies between 18 - 25 veh/mi/ln

Maximum density = 25 veh/mi/ln

Density = V​​​​​​p /S = 25

0.426V = 25 * 59.3

V = 3480 veh/hr

b) V = 5435 veh/hr

LOS dropping to F

Max density = 45 veh/mi/ln

Density = Vp/S = 45

V​​​​​​p = 45 * 59.3 = 2668.5 veh/hr/ln

V/(PHF * n * f​​​​​​HV * f​​​​​​P​​​) = 2668.5

f​​​​​​HV = 5435/(0.9*3*2668.5*1.0) = 0.754

1/(1+0.10 (E​​​​​​T -1)) = 0.754

E​​​​​​T = 4.26 ~ 3.5

<em>For 4% upgrade and 10% trucks with E​​​​​​T = 3.5, length of the grade is Greater than 1.0 miles</em>

6 0
3 years ago
Read 2 more answers
Can the United States defeat Iranian forces
Lostsunrise [7]

Answer:

yes for sure

Explanation:

iran oy has 3000 ballistic missiles but america has thousands of nukes. we can easily outnumber their forces as well. this will be a easy victory for us but will result in massive casualties on both sides

5 0
2 years ago
Read 2 more answers
A thin-walled tube with a diameter of 6 mm and length of 20 m is used to carry exhaust gas from a smoke stack to the laboratory
Molodets [167]

Answer:

Explanation:

Mean temperature is given by

T_mean = \frac{T_i + T_ \infinity}{2}\\\\T_mean = \frac{200 + 15}{2}

Tmean = (Ti + T∞)/2

T_mean = 107.5^{0}

Tmean = 107.5⁰C

Tmean = 107.5 + 273 = 380.5K

Properties of air at mean temperature

v = 24.2689 × 10⁻⁶m²/s

α = 35.024 × 10⁻⁶m²/s

\mu = 221.6 × 10⁻⁷N.s/m²

\kappa = 0.0323 W/m.K

Cp = 1012 J/kg.K

Pr = v/α  = 24.2689 × 10⁻⁶/35.024 × 10⁻⁶

              = 0.693

Reynold's number, Re

Pv = 4m/πD²

where Re = (Pv * D)/\mu

Substituting for Pv

Re = 4m/(πD\mu)

     = (4 x 0.003)/( π × 6 ×10⁻³ × 221.6 × 10⁻⁷)

     = 28728.3

Since Re > 2000, the flow is turbulent

For turbulent flows, Use

Dittus - Doeltr correlation with n = 0.03

Nu = 0.023Re⁰⁸Pr⁰³ = (h₁D)/k

(h₁ × 0.006)/0.0323 = 0.023(28728.3)⁰⁸(0.693)⁰³

(h₁ × 0.006)/0.0323 = 75.962

h₁ = (75.962 × 0.0323)/0.006

h₁ = 408.93 W/m².K

4 0
3 years ago
A cooling system load is 96,000 BTUh sensible. How much chilled air is required to satisfy the load if the system is designed fo
Natalija [7]

Answer:

For 20^{\circ} - 5.556 lb/s

For 15^{\circ} - 7.4047 lb/s

Solution:

As per the question:

System Load = 96000 Btuh

Temperature, T = 20^{\circ}

Temperature rise, T' = 15^{\circ}

Now,

The system load is taken to be at constant pressure, then:

Specific heat of air, C_{p} = 0.24 btu/lb ^{\circ}F

Now, for a rise of 20^{\circ} in temeprature:

\dot{m}C_{p}\Delta T = 96000

\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 20} = 20000 lb/h = \frac{20000}{3600} = 5.556 lb/s

Now, for 15^{\circ}:

\dot{m}C_{p}\Delta T = 96000

\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 15} = 26666.667 lb/h = \frac{26666.667}{3600} = 7.4074 lb/s

4 0
2 years ago
The entire system of components that produces power and transmits it to the road is called the vehicle's _____.
IrinaK [193]

Answer:

Powertrain

Explanation:

6 0
3 years ago
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