If the atmosphere can support a column of mercury 760 mm high at sea level, what hright of a hypothetical liquid whose density is 1.40 times the density of mercy could be supported?

To determine the height of a hypothetical liquid that could be supported by the atmosphere, we can use the concept of pressure.

1. The column of mercury with a height of 760 mm is supported by atmospheric pressure. This pressure is equivalent to 1 atmosphere (atm) or 101,325 pascals (Pa).

2. For a given liquid, the pressure exerted by the column of liquid is determined by its height and density. The greater the density of the liquid, the shorter its height needs to be to exert the same pressure as the mercury column.

3. Let's denote the height of the mercury column as h_mercury and the density of the hypothetical liquid as ρ_hypothetical.

4. We can set up a proportion to relate the heights and densities of the two liquids:

h_mercury / ρ_mercury = h_hypothetical / ρ_hypothetical

Here, ρ_mercury represents the density of mercury, and h_hypothetical is the height of the hypothetical liquid we're trying to find.

5. Given that the density of the hypothetical liquid is 1.40 times the density of mercury (ρ_hypothetical = 1.40 * ρ_mercury), we can substitute this value into the equation:

h_mercury / ρ_mercury = h_hypothetical / (1.40 * ρ_mercury)

Simplifying the equation, we have:

h_mercury = h_hypothetical / 1.40

6. Rearranging the equation to solve for h_hypothetical, we get:

h_hypothetical = h_mercury * 1.40

7. Plugging in the given value for h_mercury (760 mm), we can calculate the height of the hypothetical liquid:

h_hypothetical = 760 mm * 1.40
= 1,064 mm

Therefore, a hypothetical liquid with a density 1.40 times that of mercury could be supported to a height of 1,064 mm (or 1.064 meters) by the atmosphere.

To find the height of a hypothetical liquid that can be supported by the atmosphere, we need to consider the concept of atmospheric pressure and the density of the liquid.

The atmospheric pressure at sea level can support a column of mercury 760 mm high, which is also known as 1 atmosphere (atm).

First, let's calculate the pressure exerted by the column of mercury using the equation:

Pressure = Density * gravitational acceleration * height

The density of mercury is approximately 13.6 g/cm^3, and the gravitational acceleration is approximately 9.8 m/s^2.

Converting the height from millimeters to meters:
Height = 760 mm * (1 cm / 10 mm) * (1 m / 100 cm) = 0.76 m

Now we can calculate the pressure:

Pressure = 13.6 g/cm^3 * 9.8 m/s^2 * 0.76 m
Pressure ≈ 101325 Pa (Pascal)

Since atmospheric pressure is approximately 101325 Pa, the pressure exerted by the column of mercury is equal to 1 atm.

To find the height of the hypothetical liquid that can be supported, we can use the same formula by rearranging it:

Height = Pressure / (Density * gravitational acceleration)

Let's assume the density of the hypothetical liquid is 1.40 times the density of mercury (13.6 g/cm^3).

Density of the hypothetical liquid = 1.40 * 13.6 g/cm^3

Now we can calculate the height:

Height = (101325 Pa) / [(1.40 * 13.6 g/cm^3) * 9.8 m/s^2]
Height ≈ 5875 mm

Therefore, a hypothetical liquid with a density 1.40 times that of mercury could be supported to a height of approximately 5875 mm by the atmosphere at sea level.