An unknown liquid fills a square container of sides 10 m. What is the density of a material if the pressure 5 m below the surface is 1.6 atm? Assume g=10 m/s^2 and the pressure at the surface of the liquid is 1 atm. Also assume that 1 atm = 10^5 Pa. (Hint: make sure to use the reference surface pressure in solving the problem).
1.5 kg/m^3
Well, if we want to find the density of the liquid, we can use the equation for pressure in a fluid: pressure = density * gravity * depth.
So, we can rearrange the equation to solve for density: density = pressure / (gravity * depth).
Now, we're given that the pressure 5 meters below the surface is 1.6 atm. But since we need to use the reference surface pressure, we need to add 1 atm to that: 1.6 atm + 1 atm = 2.6 atm.
But we need to convert atm to Pa, so we can multiply by 10^5: 2.6 atm * 10^5 Pa/atm = 2.6 * 10^5 Pa.
Now, we can plug this value into our density equation, along with the given values for gravity (10 m/s^2) and the depth (5 m):
density = (2.6 * 10^5 Pa) / (10 m/s^2 * 5 m)
density = 52000 kg/m^3
So, the density of the unknown liquid is 52000 kg/m^3. Now, that's a dense liquid! Maybe it's made of jokes that are just too heavy to be funny.
To find the density of the material, we can use the hydrostatic pressure equation:
P = P0 + ρgh
where:
P is the pressure at a certain depth,
P0 is the pressure at the surface of the liquid,
ρ is the density of the liquid,
g is the acceleration due to gravity, and
h is the depth below the surface.
In this case, the pressure at a depth of 5 m below the surface is given as 1.6 atm.
First, let's convert the pressure to Pascals (Pa):
1 atm = 10^5 Pa
So, 1.6 atm = 1.6 * 10^5 Pa.
Now, we can calculate the pressure at the surface of the liquid, P0:
P0 = 1 atm = 1 * 10^5 Pa.
Next, we can substitute the values into the hydrostatic pressure equation:
P = P0 + ρgh
1.6 * 10^5 Pa = 1 * 10^5 Pa + ρ * 10 m/s^2 * 5 m
Now, let's solve for ρ:
1.6 * 10^5 Pa - 1 * 10^5 Pa = ρ * 10 m/s^2 * 5 m
0.6 * 10^5 Pa = ρ * 50 m^2/s^2
ρ = (0.6 * 10^5 Pa) / (50 m^2/s^2)
Now, let's simplify the units:
ρ = 0.12 * 10^5 Pa / (m^2/s^2)
Since 1 Pa is equivalent to 1 kg/m^2/s^2, we can simplify further:
ρ = 0.12 * 10^5 kg/m^2/s^2
Finally, we can express the density in a more commonly used unit:
ρ = 1200 kg/m^3
Therefore, the density of the material is 1200 kg/m^3.
To find the density of the material, we can use the hydrostatic pressure formula:
P = P0 + ρgh
Where:
P is the pressure at a certain depth,
P0 is the reference pressure (in this case, the pressure at the surface),
ρ is the density of the material,
g is the acceleration due to gravity, and
h is the depth below the surface.
In this problem, we know the following values:
P = 1.6 atm (given)
P0 = 1 atm (given)
g = 10 m/s^2 (given)
h = 5 m (given)
First, let's convert the pressures from atm to Pa since the units of the other variables are in meters:
1 atm = 10^5 Pa
So, the pressure at 5m below the surface is:
P = 1.6 atm × 10^5 Pa / 1 atm = 1.6 × 10^5 Pa
Next, substituting the known values into the hydrostatic pressure formula, we can solve for ρ (density):
1.6 × 10^5 Pa = 1 × 10^5 Pa + ρ × 10 m/s^2 × 5 m
Simplifying the equation:
1.6 × 10^5 Pa - 1 × 10^5 Pa = ρ × 10 m/s^2 × 5 m
0.6 × 10^5 Pa = ρ × 50 m^2/s^2
Simplifying further:
ρ = 0.6 × 10^5 Pa / (50 m^2/s^2)
ρ = 1.2 kg/m^3
Therefore, the density of the material is 1.2 kg/m^3.