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kramer
4 years ago
6

What represents a new approach to middleware by packaging together commonly used functionality. such as providingprebuilt links

to popular enterprise applications. which reduces the time necessary to develop solutions that integrateapplications from multiple vendors?A. MiddlewareB. Enterprise application integration middlewareC. Automated business processD. Ebusiness infrastructure
Physics
1 answer:
JulijaS [17]4 years ago
4 0

Answer:

B. Enterprise application integration middleware.

Explanation:

Enterprise application integration

is related to middleware technologies. Other developing EAI technologies involve Web service integration, service-oriented architecture, content integration and business processes

Enterprise application integration (EAI) is explained to be the use of technologies and services across an enterprise to enable the integration of software applications and hardware systems. Many proprietary and open projects provide EAI solution support.

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A record of travel along a straight path is as follows: 1. Start from rest with constant acceleration of 2.45 m/s^2 for 20.0 s.
Anton [14]

Answer:

a) The total displacement of the trip was 5.32 × 10³ m

b) The average speeds were:

leg 1: 24.5 m/s

leg 2: 49 m/s

leg 3: 23.9 m/s

Complete trip: 43.8 m/s

Explanation:

The position and velocity equations for an object moving along a straight line are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

If the velocity is constant, then a = 0 and x = x0 + v · t where "v" is the velocity.

a) To calculate the total displacement of the trip, let´s calculate the distance traveled in each phase.

Phase 1:

x = x0 + v0 · t + 1/2 · a · t²

x = 0 m + 0 m/s · t + 1/2 · 2.45 m/s² · (20.0 s)²

x = 490 m

The velocity reached in that phase is:

v = v0 + a · t

v = 0 m/s +  2.45 m/s² · 20.0 s

v = 49.0 m/s

Phase 2:

x = x0 + v · t

x = 490 m + 49.0 m/s · 96.0 s

x = 5.19 × 10³ m

Phase 3:

x = x0 + v0 · t + 1/2 · a · t²

x =  5.19 × 10³ m + 49 m/s · 5.44 s - 1/2 · 9.00 m/s² · (5.44 s)²

x = 5.32 × 10³ m

The total displacement of the trip was 5.32 × 10³ m

b) The average speed is calculated as the traveled distance divided by the elapsed time:

average speed v = final position - initial position / (final time- initial time)

Phase 1:

v = 490 m - 0 m / 20.0 s = 24.5 m/s

Phase 2:

v = 5.19 × 10³ m - 490 / 96.0 s

v = 48.9 m/s   (without rounding the final position the result is 49.0 m/s)

Phase 3:

v =  5.32 × 10³ m -  5.19 × 10³ m / 5.44 s = 23.9 m/s

For the complete trip:

v =  5.32 × 10³ m  - 0 m / (20.0 s + 96.0 s + 5.44 s)

v = 43.8 m/s

7 0
4 years ago
A 60.00-cm guitar string under a tension of 50.000 N has a mass per unit length of 0.100 00 g/cm chegg
Evgen [1.6K]

Fundamental frequency,

f=v2l=T/μ−−−−√2l

=(50)/0.1×10−3/10−22×0.6−−−−−−−−−−−−−−−−−−−√

=58.96Hz

Let, n th harmonic is the hightest frequency, then

(58.93)n = 20000

∴N=339.38

Hence, 339 is the highest frequency.

∴fmax=(339)(58.93)Hz=19977Hz.

<h3>What is frequency?</h3>

In physics, frequency is the number of waves that pass a given point in a unit of time as well as the number of cycles or vibrations that a body in periodic motion experiences in a unit of time. After moving through a sequence of situations or locations and then returning to its initial position, a body in periodic motion is said to have experienced one cycle or one vibration. See also simple harmonic motion and angular velocity.

learn more about frequency refer:

brainly.com/question/254161

#SPJ4

7 0
2 years ago
A 1,383 kg purple car is driving southbound on a road and collides with a 1,827 kg orange car, that was traveling 31.87 m/s east
user100 [1]

Answer:

Explanation:

We shall apply work energy theorem to calculate the initial velocity just after the collision .

Their kinetic energy will be equal to work done by friction .

force of friction = μ mg , where μ is coefficient of friction , m is total mass and g is acceleration due to gravity

force = .463 x 3210 x 9.8

= 14565.05 N

work done = force x displacement

= 14565.05 x 14.54 = 211775.88 J

now applying work energy theorem

1/2 m v² = 211775.88 , m is composite mass , v is velocity just after the collision

.5 x 3210 x v² = 211775.88

v² = 131.94

v  11.48 m /s

3 0
3 years ago
Glaciers, weather, and water shape Earth by _____.
Luden [163]
Glaciers, weather, and water shape Earth by (B.) the erosion.

Hope it helped you!

Have a great day!

-Charlie
6 0
3 years ago
Read 2 more answers
The lens-makers’ equation applies to a lens immersed in a liquid if n in the equation is replaced by n2/n1. Here n2 refers to th
pickupchik [31]

Answer:

a

The focal length of the lens in water is  f_{water} = 262.68 cm

b

The focal length of the mirror in water is  f =79.0cm

Explanation:

From the question we are told that

    The index of refraction of the lens material = n_2

    The index of refraction of the medium surrounding the lens = n_1

 

The lens maker's formula is mathematically represented as

            \frac{1}{f} = (n -1) [\frac{1}{R_1} - \frac{1}{R_2}  ]

Where f is the focal length

            n is the index of refraction

            R_1 and R_2 are the radius of curvature of sphere 1 and 2 of the lens

From the question When the lens in air  we have  

           \frac{1}{f_{air}} = (n-1) [\frac{1}{R_1} - \frac{1}{R_2}  ]

    When immersed in liquid the formula becomes

          \frac{1}{f_{water}} = [\frac{n_2}{n_1} - 1 ] [\frac{1}{R_1} - \frac{1}{R_2}  ]

The ratio of the focal length of the the two medium is mathematically evaluated as

           \frac{f_water}{f_{air}} = \frac{n_2 -1}{[\frac{n_2}{n_1} - 1] }

From the question

      f_{air }= 79.0 cm

       n_2 = 1.55

and the refractive index of water(material surrounding the lens) has a constant value of  n_1 = 1.33

         \frac{f_{water}}{79}  = \frac{1.55- 1}{\frac{1.55}{1.44}  -1}

           f_{water} = 262.68 cm

b

The focal length of a mirror is dependent on the concept of reflection which is not affected by medium around it.

   

7 0
3 years ago
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