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Sergeu [11.5K]
3 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]3 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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Maslowich

Answer:

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5 0
3 years ago
Light travels 186 000 miles per second.how many miles dose light travel in one year
Troyanec [42]

       (186,000 mi/sec) x (3,600 sec/hr) x (24 hr/da) x (365 da/yr)

  =   (186,000 x 3,600 x 24 x 365)  mi/yr

  =      5,865,696,000,000  miles per year  (rounded to the nearest million miles)
8 0
3 years ago
When work is done and a force is transferred which choice describes the movements of the object?
Arisa [49]
When work is done and a force is transferred an object must move
3 0
4 years ago
Find the moment of inertia about each of the following axes for a rod that is 0.360 {cm} in diameter and 1.70 {m} long, with a m
mamaluj [8]

The complete question is;

Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 × 10 ^(−2) kg.

A) About an axis perpendicular to the rod and passing through its center in kg.m²

B) About an axis perpendicular to the rod and passing through one end in kg.m²

C) About an axis along the length of the rod in kg.m²

Answer:

A) I = 0.012 kg.m²

B) I = 0.048 kg.m²

C) I = 8.1 × 10^(-8) kg.m²

Explanation:

We are given;

Diameter = 0.36 cm = 0.36 × 10^(−2) m

Length; L = 1.7m

Mass;m = 5 × 10^(−2) kg

A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

I = mL²/12

I = (5 × 10^(−2) × 1.7²)/12

I = 0.012 kg.m²

B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

I = mL²/3

So,

I = (5 × 10^(−2) × 1.7²)/3

I = 0.048 kg.m²

C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

We have diameter = 0.36 × 10^(−2) m, thus radius;r = (0.36 × 10^(−2))/2 = 0.18 × 10^(−2) m

I = (5 × 10^(−2) × (0.18 × 10^(−2))^2)/2

I = 8.1 × 10^(-8) kg.m²

3 0
3 years ago
Suppose a woman raises a 65 N object in 2m in 4 seconds.
Novosadov [1.4K]

Answer:

\huge\boxed{\sf P.E = 130\ Joules}

\huge\boxed{\sf P = 32.5\ Watts}

Explanation:

<u>Given Data:</u>

Weight = W = 65 N

Height = h = 2 m

Time = t = 4 secs

<u>Required:</u>

Power = P = ?

Work Done in the form of Potential Energy = P.E. = ?

<u>Formula:</u>

P.E. = Wh

P = P.E. / t

<u>Solution:</u>

P.E. = (65)(2)

P.E = 130 Joules

P = P.E. / t

P = 130 / 4

P = 32.5 Watts

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807 </h3>
8 0
3 years ago
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