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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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Answer:

\dot{W_{H} } = 4244.48 Btu/h

Explanation:

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Convert to rankine, T_{H} = 70^{0}+ 460 = 530 R

Heat is extracted at 40°F i.e T_{L} = 40^{0}F  = 40 + 460 = 500 R

Calculate the coefficient of performance of the heat pump, COP

COP = \frac{T_{H} }{T_{H} - T_{L}  } \\COP = \frac{530 }{530 - 500  }\\ COP = \frac{530}{30} \\COP = 17.67

The minimum power required to run the heat pump is given by the formula:

\dot{W_{H} } = \frac{\dot{Q_{H} }}{COP} \\...............(*)

Where the heat losses from the house, \dot{Q_{H} } = 75,000 Btu/h

Substituting these values into * above

\dot{W_{H} } = \frac{75000}{17.67} \\ \dot{W_{H} } = 4244.48 Btu/h

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3 years ago
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As = 143km ÷ 3hr

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Help plzzz!!! I only have 10 minutes to turn in
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  • The mechanic did 5406 Joules of work pushing the car.

That's the energy he put into the car.  When he stops pushing, all the energy he put into the car is now the car's kinetic energy.

  • Kinetic energy = (1/2) (mass) (speed²)

And there we have it

  • The car's mass is 3,600 kg.
  • Its speed is 'v' m/s .
  • (1/2) (mass) (v²) =  5,406 Joules

(1/2) (3600 kg) (v²) = 5406 joules

1800 kg (v²) = 5406 joules

v² = (5406 joules) / (1800 kg)

v² = (5406/1800) (joules/kg)

= = = = = This section is just to work out the units of the answer:

  • v² = (5406/1800) (Newton-meter/kg)
  • v² = (5406/1800) (kg-m²/s²  /  kg)
  • v² = (5406/1800)  (m²/s²)

= = = = =

v = √(5406/1800)  m/s

<em>v = 1.733 m/s</em>

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I would think the answer is c.
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