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Gre4nikov [31]
2 years ago
8

An OTR is removing electrodes from a client who has just received iontophoresis. Within several minutes of removing the electrod

es, what should the OTR do to the skin surface that came in contact with the electrodes?
Physics
1 answer:
Sveta_85 [38]2 years ago
4 0

Answer:

wipe the area that came into contact with the electrodes with an alcohol pad

Explanation:

According to my research on the procedure for iontophoresis, I can say that based on the information provided within the question the OTR should wipe the area that came into contact with the electrodes with an alcohol pad. This is because the alcohol pad kills any bacteria that is lingering on the skin and therefore prevents infections from occurring. Especially since the iontophoresis procedure can increase skin permeability which makes it easier for infections to arise.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

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Compare and contrast the strength of the forces between two objects with a mass of 1 kg each, a charge of 10
DochEvi [55]

Answer:

Let's see the similarities between the two forces

* are proportional to the product of a magnitude, mass or charge

* They are inversely proportional to the square of the distance

* They are long-range forces since zero is not made up to an infinite distance. The gravitational force is always attractive, the electrical force can be attractive or repulsive.

The differences in them

* The electric force in much greater than the gravitational force

* The gravitational force is always attractive, the electrical force can be attractive or repulsive.

Explanation:

Let's start by calculating each force.

Gravitational force

             F =G \frac{m_1m_2}{r^2}  

let's calculate

             F = 6.67 10⁻¹¹  1  1 / 1²

             F = 6.67 10⁻¹¹ N

Electric force

             F = k \frac{q_1q_2}{r^2}  

indicates that the charge is q = 10 C

            F = 9 10⁹ 10 10 / 1²

            F = 9 10¹¹ N

Let's see the similarities between the two forces

* are proportional to the product of a magnitude, mass or charge

* They are inversely proportional to the square of the distance

* They are long-range forces since zero is not made up to an infinite distance. The gravitational force is always attractive, the electrical force can be attractive or repulsive.

The differences in them

* The electric force in much greater than the gravitational force

* The gravitational force is always attractive, the electrical force can be attractive or repulsive.

3 0
2 years ago
You are driving to the grocery store at 20 m/s. You are 110m from an intersection when the traffic light turns red. Assume that
oksano4ka [1.4K]

As we know that reaction time will be

t = 0.50 s

so the distance moved by car in reaction time

d = vt

d = 20 \times 0.50

d = 10 m

now the distance remain after that from intersection point is given by

d = 110 - 10 = 100 m

So our distance from the intersection will be 100 m when we apply brakes

now this distance should be covered till the car will stop

so here we will have

v_f = 0

v_i = 20 m/s

now from kinematics equation we will have

v_f^2 - v_i^2 = 2 a d

0 - 20^2 = 2(a)100

a = \frac{-400}{200} = -2 m/s^2

so the acceleration required by brakes is -2 m/s/s

Now total time taken to stop the car after applying brakes will be given as

v_f - v_i = at

0 - 20 = -2 (t)

t = 10 s

total time to stop the car is given as

T = 10 s + 0.5 s = 10.5 s

3 0
2 years ago
You have a pumpkin of mass m and radius r. the pumpkin has the shape of a sphere, but it is not uniform inside so you do not kno
geniusboy [140]
<span>As it is descended from a vertical height h, The lost Potential Energy = Mgh The gained Kenetic Energy = (1/2)Mv^2; The rotational KE = (1/2)Jw^2 The angular speed w = speed/ Radius = v/R So Rotational KE = (1/2)Jw^2 = (1/2)J(v/R)^2; J is moment of inertia Now Mgh = (1/2)Mv^2 + (1/2)J(v/R)^2 => 2gh/v^2 = 1 + (J/MR^2) As v = (5gh/4)^1/2, (J/MR^2) = 2gh/v^2 - 1 => (J/MR^2) = (8gh/5gh) - 1 so (J/MR^2) = 3/5 and therefore J = (3/5)MR^2.</span>
8 0
2 years ago
A stock person at the local grocery store has a job consisting of the following five segments:
vaieri [72.5K]

Answer:

B

Explanation:

Work done can be said to be positive if the applied force has a component to be in the direction of the displacement and when the angle between the applied force and displacement is positive.

In 1 and 2 work done is positive

6 0
2 years ago
Alex goes cruising on his dirt bike. He rides 700m north, 300m east, 400 m north, 600m west, 1200m south, 300m east and finally
Bess [88]

Find Displacement and Distance

displacement ...

north is 700+400+100 =1200m n

south=1200m

1200-1200=0


east is 300+300=600m

west is 600m

600-600=0

back at dtart. displ zero


distance is 700+ 300m + 400 m + 600m + 1200m + 300m + 100m  = 3600m


3 0
2 years ago
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