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nordsb [41]
2 years ago
13

A tool use to quickly develop quality database design by reusing commonly available data models which are applicable to the ente

rprises requirements and customizing the details for the application at hand
a. integrated data model
b. integrated management information system
c. business process re-engineering
d. supply chain management
Physics
1 answer:
Trava [24]2 years ago
5 0
U can be helped at flash cards
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jeka94
I think the answer is Allele am not so sure of it and sorry if my answer is wrong or didn’t help okay
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3 years ago
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What is the relationship between the frequency and the pitch of a sound?
andrezito [222]
Choice-C is the correct one.
3 0
3 years ago
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A worker assigned to the restoration of the Washington Monument is checking the condition of the stone at the very top of the mo
Svetllana [295]

Answer:

The gravitational potential energy of the nickel at the top of the monument is 8.29 J.

Explanation:

We can find the gravitational potential energy using the following  formula.

GPE=mgh

Identifying given information.

The nickel has a mass m=0.005 \,kg, and it is a the top of Washington Monument.

The Washington Monument has a height of h=555 \, ft, thus we need to find the equivalence in meters using unit conversion in  order to find the gravitational potential energy.

Converting from feet to meters.

Using the conversion factor 1 m = 3.28 ft, we have

h = 555 \, ft \times \cfrac{1 \, m}{3.28 \, ft}

That give u s

h = 169.2 \, m

Finding Gravitational Potential Energy.

We can replace the height and mass on the formula

GPE=mgh

And we get

GPE=(0.005)(9.8)(169.2) \, J

\boxed{GPE=8.29 \,J}

The gravitational potential energy of the nickel at the top of the monument is 8.29 J.

7 0
3 years ago
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A spring balance attempts to drag an object across a table, but the object does not move because of the force of friction acting
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4 0
2 years ago
The maximum speed with which you can throw a stone is about 20 m/s. Can you hit a window 45 m away horizontally and 10 m up from
Allushta [10]

Answer:

 y = 17 m

Explanation:

For this projectile launch exercise, let's write the equation of position

          x = v₀ₓ t

          y = v_{oy} t - ½ g t²

let's substitute

          45 = v₀ cos θ t

          10 = v₀ sin θ t - ½ 9.8 t²

the maximum height the ball can reach where the vertical velocity is zero

 

           v_{y} = v_{oy} - gt

           0 = v₀ sin θ - gt

           0 = v₀ sin θ - 9.8 t

Let's write our system of equations

         45 = v₀ cos θ  t

         10 = v₀ sin θ t - ½ 9.8 t²

         0 = v₀ sin θ - 9.8 t

We have a system of three equations with three unknowns for which it can be solved.

Let's use the last two

        v₀ sin θ = 9.8 t

we substitute

        10 = (9.8 t) t - ½ 9.8 t2

        10 = ½ 9.8 t2

        10 = 4.9 t2

        t = √ (10 / 4.9)

        t = 1,429 s

Now let's use the first equation and the last one

         45 = v₀ cos θ t

         0  = v₀ sin θ - 9.8 t

         9.8 t = v₀  sin θ

         45 / t = v₀ cos θ

we divide

         9.8t / (45 / t) = tan θ

          tan θ = 9.8 t² / 45

          θ = tan⁻¹ ( 9.8 t² / 45 )

          θ = tan⁻¹ (0.4447)

          θ = 24º

Now we can calculate the maximum height

         v_y² = v_{oy}^2 - 2 g y

         vy = 0

          y = v_{oy}^2 / 2g

          y = (20 sin 24)²/2 9.8

          y = 3,376 m

the other angle that gives the same result is

       θ‘= 90 - θ

       θ' = 90 -24

       θ'= 66'

for this angle the maximum height is

 

          y = v_{oy}^2 / 2g

          y = (20 sin 66)²/2 9.8

           y = 17 m

thisis the correct

6 0
3 years ago
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