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Sergeeva-Olga [200]
1 year ago
8

A patient has an order for a drug to be infused at the rate of 5 mcg/kg/min. A 500 ml bag contains 250 mg of the drug and the pa

tient weighs 185 pounds. An infusion set with a drop factor of 20 is being used. What is the flow rate in gtts/min
Physics
1 answer:
enot [183]1 year ago
3 0

The flow rate is 17gtts/min.

<h3>What is the drug infusion rate?</h3>
  • The rate of infusion (or dosing rate) in pharmacokinetics refers to the ideal rate at which a drug should be supplied to achieve a steady state of a fixed dose that has been shown to be therapeutically effective. This rate is not only the rate at which a drug is administered.
  • The infusion volume is divided into drops, which is known as a drip-rate. The Drip Rate formula is as follows: Volume (mL) times time (h) equals drip-rate. A patient must get 1,000 mL of intravenous fluids over the course of eight hours.
  • Infusion rates of 3–4 mg/kg per minute are advised by manufacturers to reduce rate-related adverse effects. Usually, the infusion lasts for several hours. Although not advised, rates exceeding 5 mg/kg per hour may be tolerated by some patients.
  • If no negative reactions occur, the rate may be increased in accordance with the table every 30 minutes up to a maximum rate of 3 ml/kg/hour (not to exceed 150 ml/hour).

To find the flow rate is 17gtts/min:

\frac{18 units}{h} \frac{100ml}{100units} \frac{500ml}{250mg} \frac{1mg}{1000mgc} \frac{20gtts}{ml} =\frac{17gtts}{min}

Therefore, The flow rate is 17gtts/min.

To learn more about infusion rate, refer to:

brainly.com/question/22761958

#SPJ9

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16

Explanation:

solution:

by taking the ratio of the energy E_n,be of the nth orbit of a beryllium atom

Z_be=4 to the energy E_n,h of the nth orbit of a hydrogen (Z_h=1) atom  gives

E_n,B/E_n,H=-(2.18*10^-18)*Z^2_BE/-(2.18*10^-18)*Zh^2/n^2

                    =Z^2_BE+/Z^2_H

                    =(4)^2/(1)^2

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What average energy of particles does temperature measures?​
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A flywheel in a motor is spinning at 530 rpm when a power failure suddenly occurs. The flywheel has mass 40.0kg and diameter 75.
aleksley [76]

Answer:

 w = 25.05 rad / s ,     α = 0.7807 rad / s² ,   θ = 1972.75

Explanation:

This is a kinematic rotation exercise, let's start by looking for the acceleration when the engine is off

            θ = w₀ t - ½ α t²

            α = (w₀t - θ) 2/t²

           

let's reduce the magnitudes to the SI system

 

w₀ = 530 rev / min (2pi rad / 1 rev) (1 min / 60 s) = 55.5 rad / s

θ = 250 rev (2pi rad / 1 rev) = 1570.8 rad

 

let's calculate the angular acceleration

           α = (55.5 39 - 1570.8) 2/39²

           α = 0.7807 rad / s²

having the acceleration we can calculate the final speed

           w = w₀ - ∝ t

           w = 55.5 - 0.7807 39

           w = 25.05 rad / s

the time to stop w = 0

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           t = wo / alpha

           t = 55.5 / 0.7807

           t = 71.09 s

         

the angle traveled

       w² = w₀⁹ - 2 α θ

       w = 0

      θ = w₀² / 2α

let's calculate

      θ = 55.5 2 / (2 0.7807)

        θ = 1972.75

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