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Дано
h=6км
p=1030кл/м³
g=
p=?-
Предмет:
Физика -
Автор:
yasminevgiu - 5 лет назад
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Ответы 1
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судя по всему здесь нужно определить гидростатическое давление столба жидкости,
Дано
h=6км = 6000м
p=1030кг/м³
g= 10 Н/ кг
p=?
Решение
р= pgh =1030кг/м³ *6000м*10 Н/ кг = 62МПа
Ответ: р = 62МПа
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Автор:
miloseo9 - 5 лет назад
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0
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ПЕРЕВЕДИТЕ! ПЛИЗ
Text 1. FUNCTIONAL UNITS
OF DIGITAL COMPUTERS
As we know, all computer operations can be grouped into five
functional categories. The method in which these five functional
categories are related to one another represents the functional
organization of a digital computer. By studying the functional
organization, a broad view of the computer is received.
The five major functional units of a digital computer are:
1) Input— to insert outside information into the machine;
2) Storage or memory — to store information and make it avail
able at the appropriate time; 3) Arithmetic-logical unit — to
perform the calculations; 4) Output — to remove data from the
machine to the outside world and 5) Control unit — to cause
all parts of a computer to act as a team.
Figure 5 shows how the five functional units of the computer act together. A complete set of instructions and data are usually fed through the input equipment to the memory where they
are stored. Each instruction is then fed to the control unit. The
control unit interprets the instructions and issues commands to
the other functional units to cause operations to be performed
on the data. Arithmetic operations are performed in the arithmetic-logical unit, and the results are then fed back to the mem-огу. Information may be fed from either the arithmetic unit or
the memory through the output equipment to the outside world.
The five units of the computer must communicate with each
other. They can do this by means of a machine language which
uses a code composed of combinations of electric pulses. These
pulse combinations are usually represented by zeros and ones,
where the one may be a pulse and the zero — a no-pulse. Numbers are communicated between one unit and another by means
of these one-zero or pulse — no-pulse combinations. The input has the additional job of converting the information fed in
by the operator into machine language. In other words, it transaltes from our language into the pulse — no-pulse combinations
understandable to the computer. The output's additional job is
converting the pulse — no-pulse combinations into a form understandable to us, such as a printed report-
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