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agasfer [191]
2 years ago
14

A radio frequency identification application would most likely interface with a (an):_________

Physics
1 answer:
Irina-Kira [14]2 years ago
4 0

A radio frequency identification application would most likely interface with an Operational Data Store.

The Operations Data Store (ODS) is a central database that provides the latest data snapshots from multiple transaction systems for operational reporting.

It allows organizations to combine data in its original format from various sources into a single destination to provide business reporting.

ODS contains integrated updates from operational sources and supports business intelligence (BI) tools to facilitate tactical decision making.

For example, an administrator can configure ODS to pull weekly batches of data from a billing application that is rarely updated, importing individual transaction records as they occur in the sales database(thanks to these database triggers), then combine the two into new relational tables.

As a result, querying and reporting on operational data in ODS comes with the assurance that these integrated tables contain the latest and most relevant snapshots of the business.

Learn more about Operations Data Store here : brainly.com/question/14925154

#SPJ4

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A particle has 3 x 10^15 eV has a charge of 3uC is placed on a certain field.
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8 0
3 years ago
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Rank the following objects by their accelerations down an incline (assume each object rolls without slipping) from least to grea
Alexxx [7]

Answer:

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

Explanation:

To answer this question, let's analyze the problem. Let's use conservation of energy

Starting point. Highest point

          Em₀ = U = m g h

Final point. To get off the ramp

          Em_f = K = ½ mv² + ½ I w²

notice that we include the kinetic energy of translation and rotation

         

energy is conserved

        Em₀ = Em_f

        mgh = ½ m v² +1/2 I w²

angular and linear velocity are related

         v = w r

         w = v / r

we substitute

          mg h = ½ v² (m + I / r²)

          v² = 2 gh   \frac{m}{m+ \frac{I}{r^2} }

          v² = 2gh    \frac{1}{1 + \frac{I}{m r^2} }

this is the velocity at the bottom of the plane ,, indicate that it stops from rest, so we can use the kinematics relationship to find the acceleration in the axis ax (parallel to the plane)

         v² = v₀² + 2 a L

where L is the length of the plane

         v² = 2 a L

         a = v² / 2L

we substitute

         a = g \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

let's use trigonometry

         sin θ = h / L

         

we substitute

         a = g sin θ   \ \frac{h}{L} \  \frac{1}{1+ \frac{I}{m r^2 } }

the moment of inertia of each object is tabulated, let's find the acceleration of each object

a) Hollow cylinder

      I = m r²

we look for the acerleracion

      a₁ = g sin θ    \frac{1}{1 + \frac{mr^2 }{m r^2 } }1/1 + mr² / mr² =

      a₁ = g sin θ    ½

b) solid cylinder

       I = ½ m r²

       a₂ = g sin θ  \frac{1}{1 + \frac{1}{2}  \frac{mr^2}{mr^2} } = g sin θ   \frac{1}{1+ \frac{1}{2} }

       a₂ = g sin θ   ⅔

c) hollow sphere

     I = 2/3 m r²

     a₃ = g sin θ   \frac{1}{1 + \frac{2}{3} }

     a₃ = g sin θ \frac{3}{5}

d) solid sphere

     I = 2/5 m r²

     a₄ = g sin θ  \frac{1 }{1 + \frac{2}{5} }

     a₄ = g sin θ  \frac{5}{7}

We already have all the accelerations, to facilitate the comparison let's place the fractions with the same denominator (the greatest common denominator is 210)

a) a₁ = g sin θ ½ = g sin θ      \frac{105}{210}

b) a₂ = g sinθ ⅔ = g sin θ     \frac{140}{210}

c) a₃ = g sin θ \frac{3}{5}= g sin θ       \frac{126}{210}

d) a₄ = g sin θ \frac{5}{7} = g sin θ      \frac{150}{210}

the order of acceleration from lower to higher is

   

     a₁ <a₃ <a₂ <a₄

acceleration are

     hollow cylinder < hollow sphere < solid cylinder < solid sphere

8 0
3 years ago
Estimate how long it took King Kong to fall straight down from the top of the Empire State Building (380 m high). Express your a
ankoles [38]

Answer:

It will take 8.80 sec to fall from the building

Explanation:

We have given height pf the state building h = 380 m

Initial velocity will be 0 m /sec

So u = 0 m/sec

Acceleration due to gravity g=9.8m/sec^2

We have to find the fall time

According to second equation of motion h=ut+\frac{1}{2}gt^2

So 380=0\times t+\frac{1}{2}\times 9.8\times t^2

t^2=77.55

t = 8.80 sec

3 0
3 years ago
Two cars approach a street corner at right angles to each
irakobra [83]

Answer:

the angle is given by

Tan theta = 35/59 = 0.59

so theta = Tan ^-1 ( 0.59 )

theta = 30.54 deg.

4 0
3 years ago
BRAINLIEST PLZ HELP!
balu736 [363]
The correct answer should be A
5 0
3 years ago
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