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2 years ago

The length of a logical record is fixed for a given operating system. a) true b) false

1 answer:

8 0

Answer:

False - An operating system should allow the user to specify the length of the physical record. If the input physical record is very large the operating system needs to know where one logical record ends and another begins.

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On a day the air is still a windmill still posses what type of energy is that ?

Montano1993 [528]

Well, it's up on top of a pole or pedestal of some sort,
so it has some gravitational potential energy relative to
the ground. In other words, if it somehow became detached
from its structure and fell to the ground, it would make quite
an energetic splat when it got there.

Also, the windmill is at the temperature of the air around it,
which is far from Absolute Zero, so the windmill holds a lot of
thermal (heat) energy.

Then I guess there's the matter of the chemical energy in the
molecules of the material that the windmill is made of, and the
nuclear energy in its atoms.

6 0

3 years ago

A sled is accelerating down a hill at a rate of 1 m s2 . If the mass of the sled is suddenly cut in half and the net force on th

mihalych1998 [28]

We have that F=ma from the 2nd Newton law where F is the force, m is the mass and a is the acceleration. Suppose we have that F' is the new force and m' is the new mass. Then, we have that a'=F'/m' still, by rearranging Newton's law. We are given that F'=2F and m'=m/2. Hence,
$a'= \frac{2F}{ \frac{m}{2} } = \frac{4F}{m} = 4\frac{F}{m}$
But now, we have from F=ma, that a=F/m and we are given that a=1m/s^2.
We can substitute thus, a'=4a=4*1m/s^2=4m/s^2.

4 0

3 years ago

The group number of on the periodic table corresponds to which characteristics of the elements of that group ?

vovangra [49]

Answer:

Explanation:

In 1789, Antoine Lavoisier published a list of 33 chemical elements. Although Lavoisier grouped the elements into gases, metals, non-metals, and earths, chemists spent the following century searching for a more precise classification scheme. In 1829, Johann Wolfgang Döbereiner observed that many of the elements could be grouped into triads (groups of three) based on their chemical properties. Lithium, sodium, and potassium, for example, were grouped together as being soft, reactive metals. Döbereiner also observed that, when arranged by atomic weight, the second member of each triad was roughly the average of the first and the third.[19] This became known as the Law of Triads.[20] German chemist Leopold Gmelin worked with this system, and by 1843 he had identified ten triads, three groups of four, and one group of five. Jean Baptiste Dumas published work in 1857 describing relationships between various groups of metals. Although various chemists were able to identify relationships between small groups of elements, they had yet to build one scheme that encompassed them all.[19]

3 0

4 years ago

a disk of a radius 50 cm rotates at a constant rate of 100 rpm. what distance in meters will a point on the outside rim travel d

trasher [3.6K]

50π m ≈ 157 m
Explanation:
100 rev/min (2π rad/rev) / (60 sec/min) = 3⅓π rad/s
d = ωrt = 3⅓π(0.50)(30) = 50π m ≈ 157 m

8 0

3 years ago

La velocidad de un tren se reduce uniformemente de 12m/s a 5m/s. Sabiendo que durante ese tiempo recorre una distancia de 100 m.

KATRIN_1 [288]

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

s = 21,0 m

Por lo tanto, la distancia que recorre hasta detenerse asumiendo la misma aceleración es 21.0 m

3 0

3 years ago