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1 year ago

To model time-variant data, one must create a new entity in an m:n relationship with the original entity.

1 answer:

6 0

To model time-variant data, one must create a new entity in an m:n relationship with the original entity, is a False statement.

Like the majority of software engineering initiatives, the ER process begins with gathering user requirements. What information must be retained, what questions must be answered, and what business rules must be implemented (For instance, if the manager column in the DEPARTMENT table is the only column, we have simply committed to having one manager for each department.)

The end result of the E-R modeling procedure is an E-R diagram that can be roughly mechanically transformed into a set of tables. Tables will represent both entities and relationships; entity tables frequently have a single primary key, but the primary key for relationship tables nearly invariably involves numerous characteristics.

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Using the graph above, explain what happened on the amount carbon in the atmosphere 1960 to 2020.

Ivenika [448]

Answer:

No graph...

Explanation:

4 0

3 years ago

Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

3 0

3 years ago

A 1.1 kg ball is attached to a ceiling by a 2.16 m long string. The height of the room is 5.97 m . The acceleration of gravity i

nydimaria [60]

1. -23.2 J

The gravitational potential energy of the ball is given by

$U=mgh$

where

m = 1.1 kg is the mass of the ball

g = 9.8 m/s^2 is the acceleration of gravity

h is the height of the ball, relative to the reference point chosen

In this part of the problem, the reference point is the ceiling. So, the ball is located 2.16 m below the ceiling: therefore, the heigth is

h = -2.16 m

And the gravitational potential energy is

$U=(1.1 kg)(9.8 m/s^2)(-2.16 m)=-23.2 J$

2. 41.1 J

Again, the gravitational potential energy of the ball is given by

$U=mgh$

In this part of the problem, the reference point is the floor.

The height of the ball relative to the floor is equal to the height of the floor minus the length of the string:

h = 5.97 m - 2.16 m = 3.81 m

And so the gravitational potential energy of the ball relative to the floor is

$U=(1.1 kg)(9.8 m/s^2)(3.81 m)=41.1 J$

3. 0 J

As before, the gravitational potential energy of the ball is given by

$U=mgh$

Here the reference point is a point at the same elevation of the ball.

This means that the heigth of the ball relative to that point is zero:

h = 0 m

And so the gravitational potential energy is

$U=(1.1 kg)(9.8 m/s^2)(0 m)=0 J$

4 0

3 years ago

a car is moving 8.80 m/s when it begins to accelerate at 2.45 m/s^2. how much time does it take to trav 138m. please help me (':

Ulleksa [173]

Answer:

7.6 s

Explanation:

Considering kinematics formula for final velocity as

$v^{2}=u^{2}+2as$

Where v and u are final and initial velocities, a is acceleration and s is distance moved.

Making v the subject then

$v=\sqrt{u^{2}+2as}$

Substituting 8.8 m/s for u, 138 m for s and 2.45 m/s2 for a then

$v=\sqrt{8.8^{2}+2*2.45*138}\\v=27.45 m/s$

Also, v=u+at and making t the subject of the formula

$t=\frac {v-u}{a}$

Substituting 27.45 m/s for v, 8.8 m/s for u and 2.45 m/s for a then

$t=\frac {27.45-8.8}{2.45}=7.6122448979591\approx 7.6s$

Therefore, it needs 7.6 seconds to travel

7 0

3 years ago

Is Mercury a heavier element than tin

Kryger [21]

Yes because mercury has more protons and electrons that tin. (30 more)

4 0

3 years ago