The SI unit of pressure, the Pascal ($Pa$), can be expressed in terms of the fundamental SI units of mass ($kg$), length ($m$), and time ($s$) as ______.
$kg \cdot m^{-1} \cdot s^{-2}$
This explanation breaks down how to express the SI unit of pressure, the Pascal ($Pa$), using the fundamental SI units: mass ($kg$), length ($m$), and time ($s$). Understanding these unit relationships is key in physics.
Pressure ($P$) is defined as the amount of force ($F$) acting perpendicularly on a unit of area ($A$). The basic formula relating these quantities is:
$$ P = \frac{F}{A} $$
To find the units of pressure in terms of fundamental units, we first need to express the unit of force, the Newton ($N$), using mass, length, and time. Newton's second law of motion provides this relationship: $F = ma$, where '$m$' is mass and '$a$' is acceleration.
Therefore, the Newton ($N$) can be expressed in fundamental SI units as:
$$ 1 \ N = 1 \ kg \cdot m \cdot s^{-2} $$
Next, we consider the unit of area ($A$). Area is a measure of two-dimensional space.
Consequently, the SI unit for area is the square meter:
$$ 1 \ m^2 $$
Now, we can determine the fundamental units of pressure by substituting the derived units of force and area into the pressure formula ($P = F/A$). The SI unit of pressure, the Pascal ($Pa$), is equivalent to one Newton per square meter ($1 \ Pa = 1 \ N/m^2$).
Expressing this using the fundamental units:
$$ Pa \ \equiv \frac{\text{Units of Force}}{\text{Units of Area}} $$
$$ Pa \ \equiv \frac{kg \cdot m \cdot s^{-2}}{m^2} $$
To simplify this expression, we combine the powers of the meter ($m$) unit. We have $m^1$ in the numerator and $m^2$ in the denominator:
$$ Pa \ \equiv kg \cdot m^{(1-2)} \cdot s^{-2} $$
This simplifies to:
$$ Pa \ \equiv kg \cdot m^{-1} \cdot s^{-2} $$
The SI unit of pressure, the Pascal ($Pa$), when expressed in terms of the fundamental SI units of mass ($kg$), length ($m$), and time ($s$), is $kg \cdot m^{-1} \cdot s^{-2}$.
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