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2 years ago

11

The analysts at Techno InfoSystems are considering the four-model approach to system development for a new client.

1 answer:

garri49 [273]2 years ago

4 0

It will provide a clear picture of current system functions before any modifications or improvements are made is a benefit if they use the four-model approach.

<u>Option: D</u>

<u>Explanation:</u>

The four model approach is followed by number of analyst which showcase that they construct the physical and logical model of both current and new system. The most important advantage of such approach is it portrays transparent image of ongoing system, before one apply any modification or variation. This is necessary because the flaws which generated earlier in system may affect the SDLC phases and outcome of such process may result into unsatisfied user by paying additional cost. This can be avoided by taking additional steps which make it worth it. The major disadvantage of such approach is the added time and cost of constructing model in both current system.

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A 0.74 mF capacitor is connected to a standard outlet (rms voltage 82 V, frequency 49 Hz ). Determine the magnitude of the curre

disa [49]

Answer:

I = 26.36 cosω t A

Explanation:

Given that

C=0.74 mF

Vrms= 82 V

Frequency ,f= 49 Hz

We know that ω = 2 π f

ω = 2 x π x 49

ω = 307.72 rad/s

As we know that voltage given as

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$V_o=\sqrt2\ V_{rms}$

$V_o=\sqrt2\ \times 82$ \

Vo=115.96 V

V=115.96 sinω t

The current given as

$I=C\dfrac{dV}{dt}$

$I=0.74\times \dfrac{dV}{dt}\ mA$

$\dfrac{dV}{dt}=115.96\omega cos\omega t$

$I=0.74\times 115.96\times 307.22 cos\omega t\ mA$

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7 0

2 years ago

A sphere with a charge q is fixed at the bottom left corner of the right triangle shown in the figure. Points P and R are at the

Alexus [3.1K]

Answer:

the final potential energy of this system is 3U0/10

Explanation:

We are given

charge at left end and another test charge at point p

Potential energy is given by = $\frac{k*Q1*Q2 }{R}$

where k is electrostatics constant = $9 *10^9$

Q1 = first charge , Q2= test charge

R= distance between charges

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U0 = k*Q1*Q2 /3 ⇒ kq1q2 = 3U0 ..............1

now the test charge moves to point R

using Pytahgoreou theorem

R(distance) = $\sqrt{8^2 + 6^2}$ = 10

New Potential energy

U1 = kq1*q2 / 10

substituting kq1q2 = 3U0 from 1

U1 = 3U0/10

So this is the final potential energy of this system.

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2 years ago