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erastova [34]
4 years ago
10

Security is a major concern now that computers play such a large part in healthcare. To address this concern, as well as others

in the modern medicine era, Congress passed _______ in 1996.
A. ARLNA

B. AHIMA

C. HIPAA

D. AHA
Physics
2 answers:
jekas [21]4 years ago
7 0
The United States passed legislation in 1996 in an effort to ensure the privacy and safeguarding of all individuals’ medical data. In August of that year, the Health Insurance Portability and Accountability Act (HIPAA)<span> was signed by President Bill Clinton

answer: </span><span>C. HIPAA </span>
Fed [463]4 years ago
6 0
The correct answer is C. HIPAA

This is the acronym for <span>Health Insurance Portability and Accountability Act. This was done to prevent illegal disclosing of data which was stored in computers and was passed in 1996 when it was signed by the then incumbent president Bill Clinton.</span>
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What means of motion
ki77a [65]
Power of movement of a person place or thing
5 0
4 years ago
The power output of a tuba is 0.35 W. At what distance is the sound
STALIN [3.7K]

The distance of the sound from the tuba is 4.82 m.

<h3>Area of the tube</h3>

The area of the tuba is calculated as follows;

I = P/A

where;

  • I is intensity of sound
  • P is power
  • A is area

A = P/I

A = 0.35 / (1.2 x 10⁻³)

A = 291.67 m²

<h3>Distance of the sound</h3>

Area = 4πr²

r = \sqrt{\frac{A}{4\pi} } \\\\r = \sqrt{\frac{291.67}{4\pi} } \\\\r = 4.82 \ m

Thus, the distance of the sound from the tuba is 4.82 m.

Learn more about intensity of sound here: brainly.com/question/4431819

8 0
3 years ago
PLEASE HELP!!!!!!!!!
Strike441 [17]
I think it’s Malleability
8 0
3 years ago
Who knows the answer?
EleoNora [17]
Itssssssss c because a goes to c
7 0
3 years ago
A 13,900 N car traveling at 40.0 km/h rounds a curve of radius 1.80 ✕ 102 m.
miss Akunina [59]

a) 0.68 m/s^2

b) 964.5 N

c) 0.069

Explanation:

a)

When an object is moving in a circular motion, the direction of its velocity is changing - therefore, it has an acceleration towards the center of the circle, called centripetal acceleration.

The magnitude of the centripetal acceleration is given by

a=\frac{v^2}{r}

where

v is the speed of the object

r is the radius of the circle

For the car in this problem:

v = 40.0 km/h = 11.1 m/s is the speed

r = 180 m is the radius of the circle

Substituting, we find the acceleration:

a=\frac{11.1^2}{180}=0.68 m/s^2

b)

The centripetal force is the force that keeps the object along its circular motion. It also acts towards the center of the circle, and it is given by

F=ma

where

m is the mass of the object

a is the centripetal acceleration

Here the weight of the car is

W=mg=13,900 N

where

g=9.8 m/s^2 is the acceleration due to gravity

So the mass is

m=\frac{W}{g}=\frac{13,900}{9.8}=1418.4 kg

Therefore, the centripetal force is

F=(1418.4)(0.68)=964.5 N

c)

In this case, the force of static friction between the tires and the road provides the required centripetal force to keep the car in circular motion. This force is given by:

F_f=\mu mg

where

\mu is the coefficient of friction

Equating the frictional force to the centripetal force,

\mu mg=ma

So we get:

\mu=\frac{a}{g}

And substitutng:

a=0.68 m/s^2 (centripetal acceleration)

g=9.8 m/s^2

We find:

\mu=\frac{0.68}{9.8}=0.069

4 0
3 years ago
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