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Alinara [238K]
2 years ago
15

A physician orders Humulin R 44 units and Humulin N 40 units qam and Humulin R 35 units ac evening meal subcutaneously. How many

total units of insulin are administered each morning?
Physics
1 answer:
jekas [21]2 years ago
5 0

The question is incomplete, the concentration of qam and humulin is not given unless R is used concentration

Complete question:

A physician orders Humulin 50/50 44 units and Humulin N 40 units qam and Humulin R 35 units ac evening meal subcutaneously. How many total units of insulin are administered each morning?

Answer:

the total units of insulin admistered each morning

= 22 units of qam and humulin

Explanation:

given

44 units and Humnlin N

with concentration 50/100 = 1/2 = 0.5

∴ 44 × 0.5 ≈ 22 units in the morning

regular insulin administered each day

(22 + 35)units of qam and humulin

= 57units

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The desperate contestants on a TV survival show are very hungry. The only food they can see is some fruit hanging on a branch hi
emmasim [6.3K]

Answer:

(a) v = 15m/a

(b) No they won't feast because the rock can only rise to a height of 11.5m which is less than 15m.

Explanation:

Please see the attachment below for film solution.

6 0
1 year ago
The water level in a tank is 20 m above the ground. a hose is connected to the bottom of the tank, and the nozzle at the end of
Damm [24]

Answer:

P_(pump) = 98,000 Pa

Explanation:

We are given;

h2 = 30m

h1 = 20m

Density; ρ = 1000 kg/m³

First of all, we know that the sum of the pressures in the tank and the pump is equal to that of the Nozzle,

Thus, it can be expressed as;

P_(tank)+ P_(pump) = P_(nozzle)

Now, the pressure would be given by;

P = ρgh

So,

ρgh_1 + P_(pump) = ρgh_2

Thus,

P_(pump) = ρg(h_2 - h_1)

Plugging in the relevant values to obtain;

P_(pump) = 1000•9.8(30 - 20)

P_(pump) = 98,000 Pa

5 0
1 year ago
When a test charge q0 = 2 nC is placed at the origin, it experiences a force of 8 times 10-4 N in the positive y direction. What
ser-zykov [4K]

Answer:

Electric field, E=4\times 10^5\ N/C

Explanation:

It is given that,

Magnitude of charge, q_o=2\ nC=2\times 10^{-9}\ C

Force experienced, F=8\times 10^{-4}\ N

We need to find the electric field at the origin. It is given by :

F=q_o\times E

E=\dfrac{F}{q_o}

E=\dfrac{8\times 10^{-4}}{2\times 10^{-9}}

E=4\times 10^5\ N/C

So, the electric field at the origin is 4\times 10^5\ N/C. Hence, this is the required solution.

3 0
2 years ago
A charge q = 3 × 10-6 C of mass m = 2 × 10-6 kg, and speed v = 5 × 106 m/s enters a uniform magnetic field. The mass experiences
NeX [460]

Answer:

Magnetic field, B = 0.004 mT

Explanation:

It is given that,

Charge, q=3\times 10^{-6}\ C

Mass of charge particle, m=2\times 10^{-6}\ C

Speed, v=5\times 10^{6}\ m/s

Acceleration, a=3\times 10^{4}\ m/s^2

We need to find the minimum magnetic field that would produce such an acceleration. So,

ma=qvB\ sin\theta

For minimum magnetic field,

ma=qvB

B=\dfrac{ma}{qv}

B=\dfrac{2\times 10^{-6}\ C\times 3\times 10^{4}\ m/s^2}{3\times 10^{-6}\ C\times 5\times 10^{6}\ m/s}

B = 0.004 T

or

B = 4 mT

So, the magnetic field produce such an acceleration at 4 mT. Hence, this is the required solution.

4 0
2 years ago
A rescue helicopter wants to drop a package of supplies to isolated mountain climbers on a rocky ridge 200 m below. If the helic
andreev551 [17]

Answer:

a) 447.21m

b) -62.99 m/s

c)94.17 m/s

Explanation:

This situation we can divide in 2 parts:

⇒ Vertical : y =-200 m

y =1/2 at²

-200 = 1/2 *(-9.81)*t²

t= 6.388766 s

⇒Horizontal: Vx = Δx/Δt

Δx = 70 * 6.388766 = 447.21 m

b) ⇒ Horizontal

Vx = Δx/Δt ⇒ 70 = 400 /Δt

Δt= 5.7142857 s

⇒ Vertical:

y = v0t + 1/2 at²

-200 = v(5.7142857) + 1/2 *(-9.81) * 5.7142857²

v0= -7 m/s  ⇒ it's negative because it goes down.

v= v0 +at

v= -7 + (-9.81) * 5.7142857

v= -62.99 m/s

c) √(70² + 62.99²) = 94.17 m/s

8 0
2 years ago
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