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Sergeu [11.5K]
4 years ago
15

"You are gowning using the closed cuff method. With the assistance of the circulator, you slid your arms into the gown sleeves t

he full distance they should go. Where should your fingertips be now in relation to the gown cuff?"
Physics
1 answer:
Nastasia [14]4 years ago
4 0

Answer:

The fingertips should be at point of attachment of the cuff to the sleeve

Explanation:

In the closed cuff method, the gown is picked up from the wrapper by holding onto the exposed inside top layer.

The picked up gown should be handled by holding the region close to the neck of the gown, at the same time avoiding contact of the gown with your body or other objects that are unsterile

The arms are then slid into the gown sleeves with the hands at the shoulder level to avoid contact with the body of the gown

The arms are then further slid into the sleeves with the assistance of the circulator to the point where your fingertips are at the midpoint of the attachment of cuff and sleeve

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Why does salt dissolve in water act as an antifreeze?
aalyn [17]

Answer:

A

Explanation:

An antifreeze is a substance which when added to a liquid reduces the freezing point of the liquid. Hence the solution can only freeze at a lower temperature compared to the solvent. Examples of antifreeze includes salt and propylene glycol. Antifreeze are used in low temperature environments to prevent water from freezing quicky. Hence the answer is A.

Have a great day Aylabailey4930

8 0
3 years ago
Read 2 more answers
The velocity of a 1.3 kg remote-controlled car is plotted on the graph. The work of segment A is J.
tamaranim1 [39]

Answer: 585 J

Explanation:

We can calculate the work done during segment A by using the work-energy theorem, which states that the work done is equal to the gain in kinetic energy of the object:

W=K_f -K_i

where Kf is the final kinetic energy and Ki the initial kinetic energy. The initial kinetic energy is zero (because the initial velocity is 0), while the final kinetic energy is

K_f =\frac{1}{2}mv^2

The mass is m=1.3 kg, while the final velocity is v=30 m/s, so the work done is:

W=K_f = \frac{1}{2}(1.3 kg)(30 m/s)^2=585 J

5 0
4 years ago
Read 2 more answers
In Lesson 20, a magnesium strip was used to ignite the thermite reaction. When magnesium is placed in a flame from a small blow
nadezda [96]

Answer:

2 electrons will be needed by unbound oxygen in order to fill its 2nd shell.

Explanation:

The chemical reaction between magnesium and oxygen gives magnesium oxide as a product.The reaction is chemically represented as:

2Mg(s)+O_2(g)\rightarrow 2MgO(s)

Magnesium is a metal of group-2 with 2 valence electrons.It has atomic number of 12.

[Mg]=1s^22s^22p^63s^2

In order to attain noble gas configuration it will loose two electrons.

[Mg]^{2+}=1s^22s^22p^6

Mg\rightarrow Mg^{2+}+2e^-...[1]

Oxygen is a non metal of group-16 with 6 valence electrons..It has atomic number of 8.

[O]=1s^22s^22p^4

In order to attain noble gas configuration it will gain two electrons.

[O]^{2-}=1s^22s^22p^6

O+2e^-\rightarrow O^{2-}..[2]

2 electrons will be needed by unbound oxygen in order to fill its 2nd shell.

6 0
4 years ago
Shanti is riding on a train that is moving at a speed of 90 km/h. He is carrying a power cord for his phone that is 1.2 m long.
faltersainse [42]

Answer:equal to 1.2 m

Explanation: i took the test

3 0
3 years ago
Read 2 more answers
A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At
yaroslaw [1]

The electric field of a very large (essentially infinitely large) plane of charge is given by:

E = σ/(2ε₀)

E is the electric field, σ is the surface charge density, and ε₀ is the electric constant.

To determine σ:

σ = Q/A

Where Q is the total charge of the sheet and A is the sheet's area. The sheet is a square with a side length d, so A = d²:

σ = Q/d²

Make this substitution in the equation for E:

E = Q/(2ε₀d²)

We see that E is inversely proportional to the square of d:

E ∝ 1/d²

The electric field at P has some magnitude E. Now we double the side length of the sheet while keeping the same amount of charge Q distributed over the sheet. By the relationship of E with d, the electric field at P must now have a quarter of its original magnitude:

E_{new} = E/4

4 0
3 years ago
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