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
1 answer:
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:
Therefore, The flow rate is 17gtts/min.
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Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y = t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45 )
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² = - 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct
We use the equation of motion for vertical component,
Here, is displacement of bullet,
is vertical initial velocity of bullet which is equal to zero because bullet was fired horizontally, and t is time of flight.
Therefore,
Given,
Substituting the values, we get time of flight