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Keith_Richards [23]
3 years ago
7

Infuse 1000 mL of NS. The infusion set delivers 15 gtt/mL. Give 300 mL bolus over 20 minutes and then infuse at the hourly rate

of 60 mL/hr. What should the rate of infusion (in mL/hr) be set after the bolus dose?
Physics
1 answer:
Bas_tet [7]3 years ago
4 0

Answer:

rate of infusion is 900 mL/hr

Explanation:

given data

Infuse I1 = 1000 mL

delivers = 15 gtt/mL

Infuse I2 = 300 mL

time t= 20 min

rate = 60 mL/hr

to find out

rate of infusion

solution

we know here we give 300 mL infuse in 20 min

so here for 20 min

rate of infusion is express as

rate of infusion = I2 / t

rate of infusion = 300 / 20

rate of infusion = 15 mL / min

rate of infusion = 15 × 60 = 900

so rate of infusion is 900 mL/hr

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Answer:

ω2  =  216.47 rad/s

Explanation:

given data

radius r1 =  460 mm

radius r2 = 46 mm

ω =  32k rad/s

solution

we know here that power generated by roller that  is

power = T. ω    ..............1

power = F × r × ω

and this force of roller on cylinder is equal and opposite force apply by roller

so power transfer equal in every cylinder so

( F × r1 × ω1)  ÷ 2 = (  F × r2 × ω2 )  ÷  2    ................2

so

ω2  =  \frac{460\times 32}{34\times 2}

ω2  =  216.47

8 0
3 years ago
Consider a uniform hoop of radius r and mass m rolling without slipping. which is larger, its translational kinetic energy or it
OlgaM077 [116]
<span>translational kinetic energy is larger than its rotational kinetic energy</span>
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3 years ago
50g of ice at 0°C is mixed with 50g of water at 80°C, what will be the final temperature of a mixture in
xxTIMURxx [149]

Answer:

0° C

Explanation:

Given that

Mass of ice, m = 50g

Mass of water, m(w) = 50g

Temperature of ice, T(i) = 0° C

Temperature of water, T(w) = 80° C

Also, it is known that

Specific heat of water, c = 1 cal/g/°C

Latent heat of ice, L(w) = 89 cal/g

Let us assume T to be the final temperature of mixture.

This makes the energy balance equation:

Heat gained by ice to change itself into water + heat gained by melted ice(water) to raise its temperature at T° C = heat lost by water to reach at T° C

m(i).L(i) + m(i).c(w)[T - 0] = m(w).c(w)[80 - T], on substituting, we have

50 * 80 + 50 * 1(T - 0) = 50 * 1(80 - T)

4000 + 50T = 4000 - 50T

0 = 100 T

T = 0° C

Thus, the final temperature is 0° C

3 0
3 years ago
A truck moves 70 m east, then moves 120 m west, and finally moves east again a distance of 90 m. If east is chosen as the positi
asambeis [7]

Answer:

140m east

Explanation:

If East is positive then lets rephrase the problem into integers

A truck moves +70 m, then moves -120m, and finally moves +90m.

So totally Displacement = +70-120+90= +140m

Since east is positive, the trucks resultant displacement is 140 m east of origin

6 0
3 years ago
Question 1(Multiple Choice Worth 4 points) The star named Canopus has a declination of approximately –52°. Which of these statem
Y_Kistochka [10]

Answer:

It is 52° below the celestial equator.

Explanation:

The declination is the angle in degrees measured north (+) or south (-) of the an imaginary line called the celestial equator.

The celestial equator is a projection of the earth's equator on the celestial sphere. imaginary

The star named Canopus has a declination of approximately –52°.

Since the angle is negative, this shows that it is south or below the celestial equator and at 52° south of the celestial equator.

Thus, the star named Caponus is 52° below the celestial equator.

8 0
3 years ago
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