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2 years ago

How can older adults avoid polypharmacy? what is the nurse’s role in preventing polypharmacy?.

1 answer:

3 0

The nurse's duty in avoiding polypharmacy is to keep a thorough record of all medications you take, including their brand and generic names, dosages, and intended uses.

<h3>What is polypharmacy?</h3><h3 />

Polypharmacy includes the usage of many drugs as well as possibly unsuitable pharmaceuticals.

Assist patients and families close to getting a complete list of drugs. Rearrange the patient's electronic health record prescription list.

Look out for unsuitable and inaccurate medications. When rescribing drugs, use caution.

Keeping a complete record of all drugs you take, including their brand and generic names, doses, and intended use is the nurse’s role in preventing polypharmacy.

Hence, nurses play an important role in preventing polypharmacy.

To learn more about the polypharmacy refer;

brainly.com/question/12926552

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Which type of joint function as a suture to tightly bind bones together so they dont move?

netineya [11]

Fibrous joint functions as a suture to tightly bind bones together so they do not move.

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3 years ago

Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o

polet [3.4K]

Answer:

a. charge C experiences the greatest net force, and charge B receives the smallest net force

b. ratio=9

Explanation:

<u>Electrostatic Force</u>

Two point-charges $q_1$ and $q_2$ separated a distance d will exert a force on each other of a magnitude given by the Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{r^2}$

Where k is the proportional constant of value

$k=9*10^9\ N.m^2/c^2$

The diagram provided in the question shows four identical charges (let's assume their value is Q) separated by identical distance (of value d). The force between the charges next to others is

$\displaystyle F_1=\frac{k\ Q\ Q}{d^2}$

$\displaystyle F_1=\frac{k\ Q^2}{d^2}$

The force between charges separated 2d is

$\displaystyle F_2=\frac{k\ Q^2}{(2d)^2}$

$\displaystyle F_2=\frac{k\ Q^2}{4d^2}$

And the force between the charges A and D is

$\displaystyle F_3=\frac{k\ Q^2}{(3d)^2}$

$\displaystyle F_3=\frac{k\ Q^2}{9d^2}$

Now, let's analyze each charge and the force applied to them by the others

Let's recall equally signed charges repel each other and differently signed charges attrach each other

Charge A. It receives force to the left from B and C and to the right from D

$\displaystyle F_A=-F_1-F_2+F_3=-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}$

$\displaystyle F_A=\frac{k\ Q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})$

$\displaystyle F_A=-\frac{41}{36}F_1$

Charge B. It receives force to the right from A and D and to the left from C

$\displaystyle F_B=F_1-F_1+F_2=\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{4d^2}$

$\displaystyle F_B=\frac{1}{4}F_1$

Charge C. It receives forces to the right from all charges.

$\displaystyle F_C=F_2+F_1+F_1=\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{d^2}$

$\displaystyle F_C=\frac{9}{4}F_1$

Charge D. It receives forces to the left from all charges

$\displaystyle F_D=-F_3-F_2-F_1=-\frac{k\ Q^2}{9d^2}-\frac{k\ Q^2}{4d^2}-\frac{k\ Q^2}{d^2}$

$\displaystyle F_D=-\frac{49}{36}F_1$

Comparing the magnitudes of each force is just a matter of computing the fractions

$\displaystyle \frac{41}{36}=1.13,\ \frac{1}{4}=0.25,\ \frac{9}{4}=2.25,\ \frac{49}{36}=1.36$

We can see the charge C experiences the greatest net force, and charge B receives the smallest net force

The ratio of the greatest to the smallest net force is

$\displaystyle \frac{\frac{9}{4}}{\frac{1}{4}}=9$

The greatest force is 9 times the smallest net force

7 0

3 years ago

A beam of light, with a wavelength of 4170 Å, is shined on sodium, which has a work function (binding energy) of 4.41 × 10 –19 J

Lyrx [107]

Explanation:

Given that,

Wavelength of the light, $\lambda=4170\ A=4170\times 10^{-10}\ m$

Work function of sodium, $W_o=4.41\times 10^{-19}\ J$

The kinetic energy of the ejected electron in terms of work function is given by :

$KE=h\dfrac{c}{\lambda}-W_o\\\\KE=6.63\times 10^{-34}\times \dfrac{3\times 10^8}{4170\times 10^{-10}}-4.41\times 10^{-19}\\\\KE=3.59\times 10^{-20}\ J$

The formula of kinetic energy is given by :

$KE=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2KE}{m}} \\\\v=\sqrt{\dfrac{2\times 3.59\times 10^{-20}}{9.1\times 10^{-31}}} \\\\v=2.8\times 10^5\ m/s$

Hence, this is the required solution.

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3 years ago

Which instrument is launched into the atmosphere to collect pressure, temperature, humidity, wind speed, and other data?

koban [17]

C. Radiosonde is the answer

the above mentioned is not correct

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3 years ago

The two particles are both moving to the right. Particle 1 catches up with particle 2 and collides with it. The particles stick

Travka [436]

Answer:

c. vf is greator than v2, but less than v1

Explanation:

The principle of conservation of linear momentum states that when two or more bodies act upon one another, their total momentum remains constant.

In a system of colliding bodies the total momentum of the system just before the collision is the same as the total momentum just after the collision.

Collisions in which the kinetic energy is conserved are called elastic collision.

Collisions in which the kinetic energy is not conserved are called inelastic collisions. If the two objects stick together after the collision and move with a common velocity, the collision is said to be perfectly inelastic.

<em>The above scenario is a perfectly inelastic collision. The initial velocity of particle 1 was greater than particle 2 before collision. After collision, its velocity will reduce to a final velocity vf as it transfers some of its kinetic energy to particle 2; whereas, the velocity of particle 2 will increase to a final velocity vf as it absorbs some of the kinetic energy of particle 1.</em>

Therefore,

a. vf = v2 is wrong because vf is greater than v2

b. vf is less than v2 is wrong because vf is greater than v2

c. vf is greater than v2, but less than v1 is correct.

d. vf = v1 is wrong because vf is less than v1

4 0

3 years ago