Answer:
patient receiving drug 25 MCG/minute
Explanation:
given data
infusing = 15 ml/hr
drug = 50 mg
D5W = 500 ml
to find out
How many MCG/minute
solution
we know infusing rate is 15 ml/hr = 0.25 ml/min
so 0.25 ml drug content = 50 /500 × 0.25
0.25 ml drug content = 0.025 mg
so here
rate of drug will be 0.025 mg
rate of drug = 0.025 mg = 25 ×
gm/min
rate of drug = 25 MCG/minute
so patient receiving drug 25 MCG/minute
Answer:
1. 1 s = 1 x 10⁶ μs
2. 1 g = 0.001 kg
3. 1 km = 1000 m
4. 1 mm = 1 x 10⁻³ m
5. 1 mL = 1 x 10⁻³ L
6. 1 g = 100 dg
7. 1 cm = 1 x 10⁻² m
8. 1 ms = 1 x 10⁻³ s
Explanation:
1.
1 x 10⁻⁶ s = 1 μs
(1 x 10⁻⁶ x 10⁶) s = 1 x 10⁶ μs
<u>1 s = 1 x 10⁶ μs</u>
2.
1000 g = 1 kg
1 g = 1/1000 kg
<u>1 g = 0.001 kg</u>
3.
<u>1 km = 1000 m</u>
<u></u>
4.
<u>1 mm = 1 x 10⁻³ m</u>
<u></u>
5.
<u>1 mL = 1 x 10⁻³ L</u>
<u></u>
6.
1 x 10⁻² g = 1 dg
(1 x 10⁻² x 10²) g = 1 x 10² dg
<u>1 g = 100 dg</u>
<u></u>
7.
<u>1 cm = 1 x 10⁻² m</u>
<u></u>
8.
<u>1 ms = 1 x 10⁻³ s</u>
In order to tell a river lock attendant that you wish to go through the lock, you should <span>sound one prolonged blast followed by one short blast.
You should wait about 400 feet away from the lock and wait for the flashing light signal that allows you to enter.
Also note that </span><span>commercial traffic always have the first priority in entering the locks.</span>
Answer:
<em>The amount of electric charge transported = 0.192 C</em>
Explanation:
Electric Charge: This is defined as the product of electric current and time in an electric circuit, The S.I unit of electric charge is Coulombs (C)
Q = It..................... Equation 1
Where Q = Electric charge, I = electric current, t = time.
<em>Given:</em> I = 285 mA, t = 674 milliseconds.
<em>Conversion: (i) Convert from 285 mA to A = (285/1000) A = 0.285 A</em>
<em> (ii) convert from 674 milliseconds to seconds = (674/1000) s = 0.674 s </em>
Substituting these values into equation 1
Q = 0.285 × 0.674
<em>Q = 0.192 C</em>
<em>Therefore the amount of electric charge transported = 0.192 C</em>
<em></em>
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The angular acceleration of the blade when it's switched off is (-6800 rev/min) divided by (2.8 sec) = -2,428.6 rev/(min-sec) = -40.5 rev/sec^2 .