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ira [324]
2 years 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]2 years 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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lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A
qaws [65]

2.6×10^6\:\text{m}

Explanation:

The acceleration due to gravity g is defined as

g = G\dfrac{M}{R^2}

and solving for R, we find that

R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration F_c experienced by the satellite is equal to the gravitational force F_G or

F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

v = \dfrac{2\pi r}{T}

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}

Solving for <em>M</em>, we get

M = \dfrac{4\pi^2 r^3}{GT^2}

Putting this expression back into Eqn(1), we get

R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}

\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}

\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}

\:\:\:\:= 2.6×10^6\:\text{m}

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Answer:

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A chevrolet corvette convertible can brake to a stop from a speed of 60.0 mi/h in a distance of 123 ft on a level roadway. what
pychu [463]
By definition we have that the final speed is:
 Vf² = Vo² + 2 * a * d
 Where,
 Vo: Final speed
 a: acceleration
 d: distance.
 We cleared this expression the acceleration:
 a = (Vf²-Vo²) / (2 * d)
 Substituting the values:
 a = ((0) ^ 2- (60) ^ 2) / ((2) * (123) * (1/5280))
 a = -77268 mi / h ^ 2
 its stopping distance on a roadway sloping downward at an angle of 17.0 ° is:
 First you must make a free body diagram and see the acceleration of the car:
 g = 32.2 feet / sec ^ 2
 a = -77268 (mi / h ^ 2) * (5280/1) (feet / mi) * (1/3600) ^ 2 (h / s) ^ 2
 a = -31.48 feet / sec ^ 2
 A = a + g * sin (θ) = -31.48 + 32.2 * sin17.0
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 Clearing the braking distance:
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 d = (Vf²-Vo²) / (2 * a)
 Substituting the values:
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 answer:
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