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kobusy [5.1K]
2 years ago
5

A multidimensional database (MDDB) is a type of database in which data can be viewed from multiple dimensions; commonly used wit

h ____.
Physics
1 answer:
xeze [42]2 years ago
7 0

A multidimensional database (MDDB) is a type of database in which data can be viewed from multiple dimensions; commonly used with <u>MOLAP</u>.

A multidimensional database affords the ability to swiftly procedure statistics and generates solutions quickly. MDBs let users ask questions about corporations' operations and traits.

The multidimensional databases make use of MOLAP (multidimensional online analytical processing) to get admission to its statistics. They permit the users to fast get answers to their requests by means of generating and analyzing the records as a substitute quickly. The information in multidimensional databases is stored in a records dice format.

In a  dimensional database, you have rows and columns, represented by means of X and Y. In a multidimensional database, you have got X, Y, Z, etc. depending on the number of dimensions in your statistics. Beneath is an instance of a 3-Dimensional statistics Array represented in a relational table and in 3-D.

Learn more about the multidimensional databases here: brainly.com/question/518894

#SPJ1

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A resistor is made out of a long wire having a length L. Each end of the wire is attached to a termina of a battery providing a
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Answer:

B) 2I

Explanation:

The equation that relates voltage, current and resistance is V=RI.

The equation for the resistance of a material in terms of its resistivity, length and cross-sectional area is R=\frac{\rho L}{A}

In this case, the length is divided by 2 while keeping its resistivity (since it's the same material) and area, which means the resistance gets divided by 2. Then, looking at the equation I=V/R and keeping V constant, one deduces that since the resistance now is half than before then current now must be twice as before.

This is all intuitive in fact, cuting a homogeneous resistor in half and leaving the rest of the variables constant makes twice as easy for the electrons to cross the conductor, thus twice the current (one has to know that all the variables involved behave linearly, as the equations show).

8 0
3 years ago
PLEASE HELP THIS TEST IS DUE IN TEN MINS
nekit [7.7K]

Answer:

A. A statement of how the volume of a gas is related to its temperature.

3 0
3 years ago
Read 2 more answers
Trong máy phát điện xoay chiều ba pha khi tổng điện áp tức thời của cuộn 1,2 là e1+e2=120V thì điện áp tức thời của cuộn 3 là
NARA [144]

Answer:

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Explanation:

8 0
3 years ago
2 Why <br> Force is needed​
nirvana33 [79]

Answer:

The use of force in our everyday life is very common. We use force to walk on the road, to lift the objects, to throw a cricket ball, or to move a given body by some particular speed or direction. We are very familiar with the various effects of force. We can exert pull and push.

Explanation:

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3 0
3 years ago
Salmon often jump waterfalls to reach their
natta225 [31]

Answer:

5.0 m/s

Explanation:

The horizontal motion of the salmon is uniform, so the horizontal component of the salmon's velocity is constant and it is

v_x = u cos \theta

where u is the initial speed and \theta=37.7^{\circ}. The horizontal distance travelled by the salmon is

d=v_x t = (ucos \theta)t

where d = 1.95 m and t is the time needed to reach the final point.

Re-arranging for t,

t=\frac{d}{v_x}=\frac{d}{u cos \theta} (1)

Along the vertical direction, the equation of motion is

y=h+u_y t -\frac{1}{2}gt^2

where:

y = 0.311 m is the final height reached by the salmon

h = 0 is the initial height

u_y = u sin \theta is the vertical component of the initial velocity of the salmon

g=9.81 m/s^2 is the acceleration of gravity

t is the time

Substituting t as found in eq.(1), we get the equation

y=(u sin \theta) \frac{d}{u cos \theta}- \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}

and we can solve this formula for u, the initial speed of the salmon:

y=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}\\\\u=\sqrt{\frac{gd^2}{2(dtan \theta -y)cos^2 \theta}}=\sqrt{\frac{(9.81)(1.95)^2}{2((1.95)(tan 37.7^{\circ}) -0.311)cos^2 37.7^{\circ}}}=5.0 m/s

5 0
3 years ago
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