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aniked [119]
3 years ago
15

A client is receiving an IV solution of sodium chloride 0.9% (Normal Saline) 250 ml with amiodarone (Cordarone) 1 gram at 17 ml/

hour. How many mg/minute of amiodarone is infusing? (Enter numeric value only. If rounding is required, round to the nearest tenth.)
Physics
1 answer:
tester [92]3 years ago
6 0

Answer:

1.1mg/min

Explanation:

We are given that

Volume of solution=250 ml

Mass of amiodarone=1 g

Infusion rate=17 ml/hr

We know that

1 g= 1000 mg

Ratio of 1000g:250 ml=\frac{1000}{25}=4 mg/ml

The concentration of solution=4mg/ml

Amiodarone infusing (mg/min)=\frac{infusion \;rate(ml/hr)\times concentration}{60}

Because 1 hr= 60 minute

Amiodarone infusing=1.1mg/m

Hence, 1.1 mg/min of amiodarone is infusing.

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Knowing the constant g what will the gravitational force between two masses be if the gravitational force between them is 36n an
charle [14.2K]
The gravitational force between two masses is given by:
F=G \frac{m_1 m_2}{r^2}
where
G is the gravitational constant
m1 and m2 are the two masses
r is the separation between the two masses

We see that the force is proportional to the inverse of the square of the distance: F \sim  \frac{1}{r^2}
therefore, if the distance is tripled:
r'=3r
The force decreases by a factor 1/9:
F \sim  \frac{1}{(3r)^2}= \frac{1}{9}  \frac{1}{r^2}

Since the original force was 36 N, the new force will be
F' =  \frac{1}{9} (36 N)= 4 N
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3 years ago
Vector e is 0.111m long in a 90 deg direction. Vector f is 0.233 m long in 400 deg direction. Find the direction of their vector
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Answer:

50.6

Explanation:

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3 years ago
What is sex And new born baby why connected to women's veins?
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3 0
3 years ago
A cannon is fired from the edge of a cliff, which is 60m above the sea. The cannonball's initial velocity is 88.3m/s and it is f
wel

Answer:

a. 11.29 s b. 94.72 m/s at -39.8° c. 821.57 m​

Explanation:

a. Using y - y₀ = ut - 1/2gt² where u = vertical component of velocity = v₀sinθ where v₀ = 88.3 m/s and θ = 34.5°, y₀ = + 60 m and y = water surface = 0 m, g = 9.8 m/s² and t = time it takes the cannon to reach the water surface.

So y - y₀ = ut - 1/2gt²

y - y₀ = (v₀sinθ)t - 1/2gt²

substituting the values of the variables into the equation, we have

0 - 60 = (88.3 m/s × sin34.5°)t - 1/2 × 9.8 m/s²× t²

- 60 = 50t - 4.9t²

So, 4.9t² - 50t - 60 = 0

Using the quadratic formula to find t,

t = \frac{-(-50) +/- \sqrt{(-50)^{2} - 4 X 4.9 X -60} }{2 X 4.9} \\t = \frac{50 +/- \sqrt{2500 + 1176} }{9.8} \\t = \frac{50 +/- \sqrt{3676} }{9.8} \\t = \frac{50 +/- 60.63 }{9.8} \\t = \frac{50 + 60.63 }{9.8} or t = \frac{50 - 60.63 }{9.8} \\t = \frac{110.63 }{9.8} or t = \frac{-10.63 }{9.8} \\t = 11.29 sor -1.085

Since t cannot be negative, t = 11.29 s

b. We first need to find the impact vertical velocity component. Using

v = u - gt where u = initial vertical velocity component = v₀sinθ  and t = 11.29 s and g = 9.8 m/s². So,

v = v₀sinθ - gt

= 88.3 m/s × sin34.5° - 9.8 m/s² × 11.29 s

= 50.01 m/s - 110.64 m/s

= -60.63 m/s

Since the horizontal velocity is constant u' = v₀cosθ = 88.3 m/s × cos34.5° = 72.77 m/s.

The impact velocity is thus the resultant of the horizontal velocity and final impact velocity. So, V = √(v² + u'²)

= √((-60.63 m/s)² + (72.77 m/s)²)

= √((3676 m²/s² + 5295.48 m²/s²)

= √(8971.48 m²/s²

= 94.72 m/s

The angle θ = tan⁻¹(v/u') = tan⁻¹(-60.63 m/s ÷ 72.77 m/s) = tan⁻¹(-0.8332) = -39.8°

So the impact velocity is 94.72 m/s at -39.8°

c. The horizontal distance out from the base of the cliff that the ball strikes the water is the range, R = u't = 72.77 m/s × 11.29 s = 821.57 m​

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How are some types of collisions different from others?
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There are two general types of collisions, inelastic and elastic. 
Inelastic collisions occur when two objects collide but neither of them bounce away from each other.
Collisions in which the objects do not touch each other are elastic. (Ex: Rutherford Scattering) 
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