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ira [324]
1 year ago
14

A company has been in business for 40 years. It has tens of thousands of customer addresses in three different computer systems

and written on paper in filing cabinets. Which of these people would be MOST helpful in organizing and securing this information?
A.
database administrator

B.
operations research analyst

C.
computer network support specialist

D.
computer user support specialist
Physics
1 answer:
alexira [117]1 year ago
7 0

The person in charge of information management is the database administrator.

<h3>What is a database?</h3>

The term database refers to the arranging of the data that concerns clients in relevant entry point such as computers and file cabinets. This is the way by which companies are bale to keep track of their customers.

The person that is most helpful in the management of the tens of thousands of customer addresses in three different computer systems and written on paper in filing cabinets is the database administrator.

Learn more about database:brainly.com/question/6447559

#SPJ1

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In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i
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a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is 2.7 rad/s^2

Explanation:

a)

In a uniform circular motion, the centripetal acceleration is given by

a_c=\omega^2 r

where:

\omega is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

\omega=1.7 rad/s is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity (g=9.8 m/s^2), so:

a_c=3.2 g = 3.2(9.8)=31.4 m/s^2

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m

b)

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

a=4.4 g

So, 4.4 times the acceleration due to gravity.

The total acceleration is the resultant of the centripetal acceleration (a_c) and the tangential acceleration (a_t):

a=\sqrt{a_c^2+a_t^2}

We know that:

a = 4.4g

a_c = 3.2 g

So, we can find the tangential acceleration:

a_t = \sqrt{a^2-a_c^2}=\sqrt{(4.4g)^2-(3.2g)^2}=29.6 m/s^2

The angular acceleration is related to the tangential acceleration by

\alpha = \frac{a_t}{r}

where r = 10.9 m is the length of the centrifuge. Substituting,

\alpha = \frac{29.6}{10.9}=2.7 rad/s^2

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3 years ago
The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at
denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

A)

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m  _____ eqn (1)

<u>FOR HYDROGEN:</u>

v = √(3KT)/m = 2000 m/s  _____ eqn (2)

<u>FOR OXYGEN:</u>

velocity of oxygen = √(3KT)/(mass of oxygen)  

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

<u>velocity of oxygen = (1/4)(2000 m/s) = 500 m/s</u>

B)

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

<u>K.E of Oxygen = 4000 J</u>

C)

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

<u>v = 2149.24 m/s</u>

<u></u>

8 0
3 years ago
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