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Nana76 [90]
3 years ago
10

Susan Clerk works in the admissions office at Hillcrest Medical Center, a large teaching hospital. She would like to obtain a jo

b in the HIM department entering and abstracting data from the disease and procedure indexes for various projects. The hospital offers "minicourses" on a variety of topics. Which course topics would apply to the job she hopes to obtain
Physics
1 answer:
Viefleur [7K]3 years ago
6 0

Answer:

IT and science courses

Explanation:

As Health Information Management department requires a person who should have fair knowledge about science and medical so that he/she can deal with the data extracted from diseases. He/she should also have fair knowledge about IT so that the person can manage the files and data very well. The data should be managed accurately so different IT programs help in completing this goal. IT knowledge will help to keep data safe, accurate and complete. Due to IT knowledge they can upgrade and redesign programs according to their need.

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Bowen’s reaction series illustrates relations between:
Finger [1]

C. Temperature, chemical composition and mineral structure

Explanation:

The Bowen's reaction series illustrates the relationship between temperature, chemical composition and mineral structure.

The series is made up of a continuous and discontinuous end through which magmatic composition can be understood as temperature changes.

  • The left part is the discontinuous end while the right side is the continuous series.
  • From the series, we understand that a magmatic body becomes felsic as it begins to cool to lower temperature.
  • A magma at high temperature is ultramafic and very rich in ferro-magnesian silicates which are the chief mineral composition of olivine and pyroxene. These minerals are predominantly found in mafic- ultramafic rocks. Also, we expect to find the calcic-plagioclase at high temperatures partitioned in the magma.
  • At a relatively low temperature, minerals with frame work structures begins to form . The magma is more enriched with felsic minerals and late stage crystallization occurs here.

Learn more:

Silicate minerals brainly.com/question/4772323

#learnwithBrainly

5 0
4 years ago
What is the force on a 250 kg elevator that is falling freely at 9.8 m/sec2?​
Viefleur [7K]

Answer:

Since it is falling freely, the only force on it is its weight, w.

w = m × g = 250 kg × 9.8 m/s^2 = 2450 Newton/N

6 0
3 years ago
Read 2 more answers
A particle is moving with SHM of period pie . initially it is 10 cm from The center of the motion and moving in the positive dir
Viefleur [7K]

Answer:

y = 10.44cos(2t - 0.291) cm

Explanation:

y = Acos(2πt/T + φ) = Acos(2πt/π + φ) = Acos(2t + φ)

v = y' = -2Αsin(2t + φ)

10 = Acos(2(0) + φ) = Acosφ

6 = -2Αsin(2(0) + φ) = -2Asinφ

6/10 = -2Asinφ/Acosφ = -2tanφ

tanφ = -0.3

φ = -0.291 radians

10 = Acos(-0.291)

A = 10/cos(-0.291) = 10.44

7 0
3 years ago
Please help with this problem. Thank you.​
4vir4ik [10]

Answer:

Explanation:

F = mω²R

F = 15(2π/8.5)²(7.8)

F = 63.93044788...

F = 63.9 N

answer a) is the closest. No idea how they got a value that low unless they used a poor approximation for π.

4 0
3 years ago
Two charges, X and Y, are placed along the x-axis. Charge X is +18 nC and is placed at x = 0. Charge Y is placed at a location o
Helen [10]

Answer:

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m.

Explanation:

The Coulomb force between two charges, Q_1 and Q_2, separated by a distance, d, is given

F = k\dfrac{Q_1Q_2}{r^2}

<em>k</em> is a constant.

For the charge Z to be at equilibrium, the force exerted on it by charge X must be equal and opposite to the force exerted on it by charge Y.

It is to be placed along the <em>x</em>-axis. Hence, it is on the same line as charges X and Y.

Let the charge on Z be <em>Q</em>. It is positive.

Let the distance from charge X be <em>x m.</em> Then the distance from charge Y will be (0.60 - <em>x</em>) m.

Force due to charge X

F_X = k\dfrac{18Q}{x^2}

Force due to charge Y

F_Y = k\dfrac{-27Q}{(0.60-x)^2}

Since both forces are equal and opposite,

F_X = -F_Y

k\dfrac{18Q}{x^2} = -k\dfrac{-27Q}{(0.60-x)^2}

\dfrac{2}{x^2} = \dfrac{3}{(0.60-x)^2}

2(0.60-x)^2 = 3x^2

2(0.36-1.20x+x^2) = 3x^2

0.72-2.40x+2x^2 = 3x^2

x^2+2.40x-0.72 = 0

Applying the quadratic formula,

x = \dfrac{-2.40\pm\sqrt{2.40^2 - (4)(1)(-0.72)}}{2} = \dfrac{-2.40\pm\sqrt{8.64}}{2}

x = -2.7 or x = 0.27

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m

3 0
4 years ago
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