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Artyom0805 [142]
3 years ago
8

A client with hypertension who weighs 72.4 kg is receiving an infusion of nitroprusside (Nipride) 50 mg in D5W 250 ml at 75 ml/h

our. How many mcg/kg/minute is the client receiving? (Enter numeric value only. If rounding is required, round to the nearest tenth.)
Physics
1 answer:
Mkey [24]3 years ago
8 0

To solve this problem it is necessary to simply apply the concepts related to cross-multiply and proportion between units.

Let's start first by relating the amount of dose needed to be supplied per hour, in other words,

The infusion of 250ml should be supplied at a rate of 75ml / hour, so what amount x of mg hour should be supplied with 50Mg.

\frac{x}{75ml/hour} \rightarrow \frac{50mg}{250ml}

x \rightarrow \frac{50mg*75ml/hour}{250ml}

x \rightarrow \frac{3750mg}{250hour}

x \rightarrow 15\frac{mg}{hour}

Converting to mcg units we know that 1mg is equal to 1000mcg and that 1 hour contains 60 min, therefore

x \rightarrow 15\frac{mg}{hour}

x \rightarrow 15\frac{mg}{hour}(\frac{1000mcg}{1mg})(\frac{1hour}{60min})

x \rightarrow 250mcg/min

The dose should be distributed per kilogram of the patient so if the patient weighs 72.4kg,

Dose = \frac{250mcg/min}{72.4kg}

Dose = 3.5 \frac{mcg/min}{kg}

Therefore the client will receive 3.5mcg/kg/min.

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A skier leaves the horizontal end of a ramp with a velocity of 25.0 m/s and lands 70.0 m from the base of the ramp. How high is
Valentin [98]

Answer:

<em>The end of the ramp is 38.416 m high</em>

Explanation:

<u>Horizontal Motion </u>

When an object is thrown horizontally with an initial speed v and from a height h, it follows a curved path ruled by gravity.

The maximum horizontal distance traveled by the object can be calculated as follows:

\displaystyle d=v\cdot\sqrt{\frac  {2h}{g}}

If the maximum horizontal distance is known, we can solve the above equation for h:

\displaystyle h=\frac  {d^2g}{2v^2}

The skier initiates the horizontal motion at v=25 m/s and lands at a distance d=70 m from the base of the ramp. The height is now calculated:

\displaystyle h=\frac  {70^2\cdot 9.8}{2\cdot 25^2}

\displaystyle h=\frac  {4900\cdot 9.8}{2\cdot 625}

h= 38.416 m

The end of the ramp is 38.416 m high

8 0
3 years ago
An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi
Brrunno [24]

Answer:

The explorer should travel to reach base camp to 5.02 Km at 4.28° south of due west.

Explanation:

Using trigonometric function like Sen(Ф), Cos(Ф) and Tan(Ф) we can get distance and direction that the explorer should travel to reach base camp. When we discompound the vector X = 7.8*Cos(50) = 5.01 Km y y = 7.8 * Sen (50) - 5.6 = 5.975 - 5.6 (Km) = 0.375 (Km) so that Tan (\alpha ) = \frac{0.375}{5.01} = 4.28; \alpha = Arctang (\frac{0.375}{5.01}) = 4.28 (degree) to get how far we use Pythagorean theorem so R^{2} = x^{2}+y^{2} so that R=\sqrt{0.375^{2}+5.01^{2} } =5.02 (Km)

6 0
3 years ago
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Why air pollution is a problem?
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8 0
3 years ago
From the top of a cliff, a person uses a slingshot to fire a pebble straight downward, which is the negative direction. The init
Talja [164]

Answer:

\vec{a} = -(9.8~{\rm m/s^2})\^y

Explanation:

Regardless of the initial velocity of the pebble, the acceleration of the pebble is equal to the gravitational acceleration which is equal to 9.8 m/s2 towards downwards direction.

This can be shown by Newton's Second Law. According to the law, the net force applied on an object is equal to mass times acceleration of that object.

During the downward motion, the only force acting on the pebble is the gravitational force, hence its acceleration is equal to gravitational acceleration.

8 0
3 years ago
Raindrops acquire an electric charge as they fall. Suppose a 2.4-mm-diameter drop has a charge of +18 pC, fairly typical values.
muminat

To solve this problem we will apply the concepts related to the potential, defined from the Coulomb laws for which it is defined as the product between the Coulomb constant and the load, over the distance that separates the two objects. Mathematically this is

V = \frac{kq}{r}

k = Coulomb's constant

q = Charge

r = Distance between them

q = 18 pC \rightarrow q = 1.8*10^-11 C

d = 2.4mm \rightarrow r = 1.2 mm = 1.2*10^-3 m

Replacing,

V = \frac{kq}{r}

V = \frac{ (9*10^9)*(1.8*10^{-11})}{(1.2*10^{-3})}

V = 135 V

Therefore the potential at the surface of the raindrop is 135 V

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3 years ago
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