Pressure - MCAT Chemical and Physical Foundations of Biological Systems

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Question

Ethanol has a specific gravity of 0.780. If a long barometer tube is placed in a dish of ethanol, with the top end open to the atmosphere, to what height in the tube will the ethanol rise?

Atmospheric pressure = 1.01 * 105 Pa

g = 9.81 m/s2

Density of water = 1000 kg/m3)

Tap to reveal answer

Answer

The pressure exerted by the air on the top of the ethanol must equal the fluid pressure in the column of liquid at equilibrium. Fluid pressure is the product of density, gravity, and height of the fluid.

The density of ethanol, , may be calculated by multiplying specific gravity times the reference density of water, as specific gravity is simply a ratio of a compound's density to that of water.

Using this density in the pressure equation allows us to solve for the height of the ethanol.

1.01 * $10^{5}$ Pa = left ( $780$\frac{kg}{m^{3}$$} right )left ( $9.81$\frac{m}{s^{2}$$} right )h

After calculating, h = 13.2 m

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Question

A diver in a lake rises from deep, where the gauge pressure reads , to the surface, where the gauge pressure reads . Upon reaching the surface, by what percent has the pressure experienced by the diver changed?

Tap to reveal answer

Answer

When the diver is submerged, the pressure experienced is the sum of the gauge pressure and the atmospheric pressure above the surface.

At the surface, the gauge pressure will be zero.

The percent change in pressure will be equal to the ratio of the change in pressure to the pressure while submerged.

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Question

Two coins lie at the bottom of a pool with an uneven bottom. One is under $\small 5m$ of the water, and the second is under $\small 0.5m$ of water. What is the approximate ratio of total pressure experienced by the first coin to total pressure experienced by the second?

Tap to reveal answer

Answer

Atmospheric pressure always equals $\small 1atm$ on earth, assuming normal elevation.

Gauge pressure always equals $\small \rho gh$ , where $\small \rho$ is the density of the medium, $\small g$ is gravity, and $\small h$ is the distance to the surface.

$P_$1=1atm+(1$\frac{g}{mL}$)(9.8$\frac{m}{s^2$}$)(5m)=50atm$

$P_$2=1atm+(1$\frac{g}{mL}$)(9.8$\frac{m}{s^2$}$)(0.5m)=5.9atm$

$P_1:P_2\rightarrow 50:5.9\approx 25:3$

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Question

A tank is filled to a depth of $\small 3.27m$ with water. Oil with a density of $\small $800$\frac{kg}{m^3$}$$ is added to the tank until the final depth is $\small 5.0m$ . What is the reading of the pressure gauge at the bottom of the tank?

Ignore the contribution of atmospheric pressure.

Tap to reveal answer

Answer

The pressure contribution of the liquid can be found using the equation:

$P=SGh=\rho gh$

The product of the density, gravity, and height of the column of liquid will give the pressure. Start by calculating the pressure of the water.

$P_${water}=(1000$\frac{kg}{m^3$$}$)(9.8$\frac{m}{s^2$}$)(3.27m)=32046Pa$

Then calculate the pressure of the oil.

$P_${oil}=(800$\frac{kg}{m^3$$}$)(9.8$\frac{m}{s^2$}$)(5.0m-3.27m)$

$P_${oil}=(800$\frac{kg}{m^3$$}$)(9.8$\frac{m}{s^2$}$)(1.73m)=13563Pa$

Sum the pressure of each liquid to find the total pressure at the bottom of the tank.

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Question

Ethanol has a specific gravity of 0.780. If a long barometer tube is placed in a dish of ethanol, with the top end open to the atmosphere, to what height in the tube will the ethanol rise?

Atmospheric pressure = 1.01 * 105 Pa

g = 9.81 m/s2

Density of water = 1000 kg/m3)

Tap to reveal answer

Answer

The pressure exerted by the air on the top of the ethanol must equal the fluid pressure in the column of liquid at equilibrium. Fluid pressure is the product of density, gravity, and height of the fluid.

The density of ethanol, , may be calculated by multiplying specific gravity times the reference density of water, as specific gravity is simply a ratio of a compound's density to that of water.

Using this density in the pressure equation allows us to solve for the height of the ethanol.

1.01 * $10^{5}$ Pa = left ( $780$\frac{kg}{m^{3}$$} right )left ( $9.81$\frac{m}{s^{2}$$} right )h

After calculating, h = 13.2 m

← Didn't Know|Knew It →

Question

A diver in a lake rises from deep, where the gauge pressure reads , to the surface, where the gauge pressure reads . Upon reaching the surface, by what percent has the pressure experienced by the diver changed?

Tap to reveal answer

Answer

When the diver is submerged, the pressure experienced is the sum of the gauge pressure and the atmospheric pressure above the surface.

At the surface, the gauge pressure will be zero.

The percent change in pressure will be equal to the ratio of the change in pressure to the pressure while submerged.

← Didn't Know|Knew It →

Question

Two coins lie at the bottom of a pool with an uneven bottom. One is under $\small 5m$ of the water, and the second is under $\small 0.5m$ of water. What is the approximate ratio of total pressure experienced by the first coin to total pressure experienced by the second?

Tap to reveal answer

Answer

Atmospheric pressure always equals $\small 1atm$ on earth, assuming normal elevation.

Gauge pressure always equals $\small \rho gh$ , where $\small \rho$ is the density of the medium, $\small g$ is gravity, and $\small h$ is the distance to the surface.

$P_$1=1atm+(1$\frac{g}{mL}$)(9.8$\frac{m}{s^2$}$)(5m)=50atm$

$P_$2=1atm+(1$\frac{g}{mL}$)(9.8$\frac{m}{s^2$}$)(0.5m)=5.9atm$

$P_1:P_2\rightarrow 50:5.9\approx 25:3$

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Question

A tank is filled to a depth of $\small 3.27m$ with water. Oil with a density of $\small $800$\frac{kg}{m^3$}$$ is added to the tank until the final depth is $\small 5.0m$ . What is the reading of the pressure gauge at the bottom of the tank?

Ignore the contribution of atmospheric pressure.

Tap to reveal answer

Answer

The pressure contribution of the liquid can be found using the equation:

$P=SGh=\rho gh$

The product of the density, gravity, and height of the column of liquid will give the pressure. Start by calculating the pressure of the water.

$P_${water}=(1000$\frac{kg}{m^3$$}$)(9.8$\frac{m}{s^2$}$)(3.27m)=32046Pa$

Then calculate the pressure of the oil.

$P_${oil}=(800$\frac{kg}{m^3$$}$)(9.8$\frac{m}{s^2$}$)(5.0m-3.27m)$

$P_${oil}=(800$\frac{kg}{m^3$$}$)(9.8$\frac{m}{s^2$}$)(1.73m)=13563Pa$

Sum the pressure of each liquid to find the total pressure at the bottom of the tank.

← Didn't Know|Knew It →

Question

Ethanol has a specific gravity of 0.780. If a long barometer tube is placed in a dish of ethanol, with the top end open to the atmosphere, to what height in the tube will the ethanol rise?

Atmospheric pressure = 1.01 * 105 Pa

g = 9.81 m/s2

Density of water = 1000 kg/m3)

Tap to reveal answer

Answer

The pressure exerted by the air on the top of the ethanol must equal the fluid pressure in the column of liquid at equilibrium. Fluid pressure is the product of density, gravity, and height of the fluid.

The density of ethanol, , may be calculated by multiplying specific gravity times the reference density of water, as specific gravity is simply a ratio of a compound's density to that of water.

Using this density in the pressure equation allows us to solve for the height of the ethanol.

1.01 * $10^{5}$ Pa = left ( $780$\frac{kg}{m^{3}$$} right )left ( $9.81$\frac{m}{s^{2}$$} right )h

After calculating, h = 13.2 m

← Didn't Know|Knew It →

Question

A diver in a lake rises from deep, where the gauge pressure reads , to the surface, where the gauge pressure reads . Upon reaching the surface, by what percent has the pressure experienced by the diver changed?

Tap to reveal answer

Answer

When the diver is submerged, the pressure experienced is the sum of the gauge pressure and the atmospheric pressure above the surface.

At the surface, the gauge pressure will be zero.

The percent change in pressure will be equal to the ratio of the change in pressure to the pressure while submerged.

← Didn't Know|Knew It →

Question

Two coins lie at the bottom of a pool with an uneven bottom. One is under $\small 5m$ of the water, and the second is under $\small 0.5m$ of water. What is the approximate ratio of total pressure experienced by the first coin to total pressure experienced by the second?

Tap to reveal answer

Answer

Atmospheric pressure always equals $\small 1atm$ on earth, assuming normal elevation.

Gauge pressure always equals $\small \rho gh$ , where $\small \rho$ is the density of the medium, $\small g$ is gravity, and $\small h$ is the distance to the surface.

$P_$1=1atm+(1$\frac{g}{mL}$)(9.8$\frac{m}{s^2$}$)(5m)=50atm$

$P_$2=1atm+(1$\frac{g}{mL}$)(9.8$\frac{m}{s^2$}$)(0.5m)=5.9atm$

$P_1:P_2\rightarrow 50:5.9\approx 25:3$

← Didn't Know|Knew It →

Question

A tank is filled to a depth of $\small 3.27m$ with water. Oil with a density of $\small $800$\frac{kg}{m^3$}$$ is added to the tank until the final depth is $\small 5.0m$ . What is the reading of the pressure gauge at the bottom of the tank?

Ignore the contribution of atmospheric pressure.

Tap to reveal answer

Answer

The pressure contribution of the liquid can be found using the equation:

$P=SGh=\rho gh$

The product of the density, gravity, and height of the column of liquid will give the pressure. Start by calculating the pressure of the water.

$P_${water}=(1000$\frac{kg}{m^3$$}$)(9.8$\frac{m}{s^2$}$)(3.27m)=32046Pa$

Then calculate the pressure of the oil.

$P_${oil}=(800$\frac{kg}{m^3$$}$)(9.8$\frac{m}{s^2$}$)(5.0m-3.27m)$

$P_${oil}=(800$\frac{kg}{m^3$$}$)(9.8$\frac{m}{s^2$}$)(1.73m)=13563Pa$

Sum the pressure of each liquid to find the total pressure at the bottom of the tank.

← Didn't Know|Knew It →

Question

Ethanol has a specific gravity of 0.780. If a long barometer tube is placed in a dish of ethanol, with the top end open to the atmosphere, to what height in the tube will the ethanol rise?

Atmospheric pressure = 1.01 * 105 Pa

g = 9.81 m/s2

Density of water = 1000 kg/m3)

Tap to reveal answer

Answer

The pressure exerted by the air on the top of the ethanol must equal the fluid pressure in the column of liquid at equilibrium. Fluid pressure is the product of density, gravity, and height of the fluid.

The density of ethanol, , may be calculated by multiplying specific gravity times the reference density of water, as specific gravity is simply a ratio of a compound's density to that of water.

Using this density in the pressure equation allows us to solve for the height of the ethanol.

1.01 * $10^{5}$ Pa = left ( $780$\frac{kg}{m^{3}$$} right )left ( $9.81$\frac{m}{s^{2}$$} right )h

After calculating, h = 13.2 m

← Didn't Know|Knew It →

Question

A diver in a lake rises from deep, where the gauge pressure reads , to the surface, where the gauge pressure reads . Upon reaching the surface, by what percent has the pressure experienced by the diver changed?

Tap to reveal answer

Answer

When the diver is submerged, the pressure experienced is the sum of the gauge pressure and the atmospheric pressure above the surface.

At the surface, the gauge pressure will be zero.

The percent change in pressure will be equal to the ratio of the change in pressure to the pressure while submerged.

← Didn't Know|Knew It →

Question

Two coins lie at the bottom of a pool with an uneven bottom. One is under $\small 5m$ of the water, and the second is under $\small 0.5m$ of water. What is the approximate ratio of total pressure experienced by the first coin to total pressure experienced by the second?

Tap to reveal answer

Answer

Atmospheric pressure always equals $\small 1atm$ on earth, assuming normal elevation.

Gauge pressure always equals $\small \rho gh$ , where $\small \rho$ is the density of the medium, $\small g$ is gravity, and $\small h$ is the distance to the surface.

$P_$1=1atm+(1$\frac{g}{mL}$)(9.8$\frac{m}{s^2$}$)(5m)=50atm$

$P_$2=1atm+(1$\frac{g}{mL}$)(9.8$\frac{m}{s^2$}$)(0.5m)=5.9atm$

$P_1:P_2\rightarrow 50:5.9\approx 25:3$

← Didn't Know|Knew It →

Question

A tank is filled to a depth of $\small 3.27m$ with water. Oil with a density of $\small $800$\frac{kg}{m^3$}$$ is added to the tank until the final depth is $\small 5.0m$ . What is the reading of the pressure gauge at the bottom of the tank?

Ignore the contribution of atmospheric pressure.

Tap to reveal answer

Answer

The pressure contribution of the liquid can be found using the equation:

$P=SGh=\rho gh$

The product of the density, gravity, and height of the column of liquid will give the pressure. Start by calculating the pressure of the water.

$P_${water}=(1000$\frac{kg}{m^3$$}$)(9.8$\frac{m}{s^2$}$)(3.27m)=32046Pa$

Then calculate the pressure of the oil.

$P_${oil}=(800$\frac{kg}{m^3$$}$)(9.8$\frac{m}{s^2$}$)(5.0m-3.27m)$

$P_${oil}=(800$\frac{kg}{m^3$$}$)(9.8$\frac{m}{s^2$}$)(1.73m)=13563Pa$

Sum the pressure of each liquid to find the total pressure at the bottom of the tank.

← Didn't Know|Knew It →

Question

Ethanol has a specific gravity of 0.780. If a long barometer tube is placed in a dish of ethanol, with the top end open to the atmosphere, to what height in the tube will the ethanol rise?

Atmospheric pressure = 1.01 * 105 Pa

g = 9.81 m/s2

Density of water = 1000 kg/m3)

Tap to reveal answer

Answer

The pressure exerted by the air on the top of the ethanol must equal the fluid pressure in the column of liquid at equilibrium. Fluid pressure is the product of density, gravity, and height of the fluid.

The density of ethanol, , may be calculated by multiplying specific gravity times the reference density of water, as specific gravity is simply a ratio of a compound's density to that of water.

Using this density in the pressure equation allows us to solve for the height of the ethanol.

1.01 * $10^{5}$ Pa = left ( $780$\frac{kg}{m^{3}$$} right )left ( $9.81$\frac{m}{s^{2}$$} right )h

After calculating, h = 13.2 m

← Didn't Know|Knew It →

Question

A diver in a lake rises from deep, where the gauge pressure reads , to the surface, where the gauge pressure reads . Upon reaching the surface, by what percent has the pressure experienced by the diver changed?

Tap to reveal answer

Answer

When the diver is submerged, the pressure experienced is the sum of the gauge pressure and the atmospheric pressure above the surface.

At the surface, the gauge pressure will be zero.

The percent change in pressure will be equal to the ratio of the change in pressure to the pressure while submerged.

← Didn't Know|Knew It →

Question

Two coins lie at the bottom of a pool with an uneven bottom. One is under $\small 5m$ of the water, and the second is under $\small 0.5m$ of water. What is the approximate ratio of total pressure experienced by the first coin to total pressure experienced by the second?

Tap to reveal answer

Answer

Atmospheric pressure always equals $\small 1atm$ on earth, assuming normal elevation.

Gauge pressure always equals $\small \rho gh$ , where $\small \rho$ is the density of the medium, $\small g$ is gravity, and $\small h$ is the distance to the surface.

$P_$1=1atm+(1$\frac{g}{mL}$)(9.8$\frac{m}{s^2$}$)(5m)=50atm$

$P_$2=1atm+(1$\frac{g}{mL}$)(9.8$\frac{m}{s^2$}$)(0.5m)=5.9atm$

$P_1:P_2\rightarrow 50:5.9\approx 25:3$

← Didn't Know|Knew It →

Question

A tank is filled to a depth of $\small 3.27m$ with water. Oil with a density of $\small $800$\frac{kg}{m^3$}$$ is added to the tank until the final depth is $\small 5.0m$ . What is the reading of the pressure gauge at the bottom of the tank?

Ignore the contribution of atmospheric pressure.

Tap to reveal answer

Answer

The pressure contribution of the liquid can be found using the equation:

$P=SGh=\rho gh$

The product of the density, gravity, and height of the column of liquid will give the pressure. Start by calculating the pressure of the water.

$P_${water}=(1000$\frac{kg}{m^3$$}$)(9.8$\frac{m}{s^2$}$)(3.27m)=32046Pa$

Then calculate the pressure of the oil.

$P_${oil}=(800$\frac{kg}{m^3$$}$)(9.8$\frac{m}{s^2$}$)(5.0m-3.27m)$

$P_${oil}=(800$\frac{kg}{m^3$$}$)(9.8$\frac{m}{s^2$}$)(1.73m)=13563Pa$

Sum the pressure of each liquid to find the total pressure at the bottom of the tank.

← Didn't Know|Knew It →

Knew It

Pressure - MCAT Chemical and Physical Foundations of Biological Systems

Ethanol has a specific gravity of 0.780. If a long barometer tube is placed in a dish of ethanol, with the top end open to the atmosphere, to what height in the tube will the ethanol rise? Atmospheric pressure = 1.01 * 105 Pa g = 9.81 m/s2 Density of water = 1000 kg/m3)

A diver in a lake rises from deep, where the gauge pressure reads , to the surface, where the gauge pressure reads . Upon reaching the surface, by what percent has the pressure experienced by the diver changed?

Two coins lie at the bottom of a pool with an uneven bottom. One is under of the water, and the second is under of water. What is the approximate ratio of total pressure experienced by the first coin to total pressure experienced by the second?

Atmospheric pressure always equals on earth, assuming normal elevation. Gauge pressure always equals , where is the density of the medium, is gravity, and is the distance to the surface.

A tank is filled to a depth of with water. Oil with a density of is added to the tank until the final depth is . What is the reading of the pressure gauge at the bottom of the tank? Ignore the contribution of atmospheric pressure.

Ethanol has a specific gravity of 0.780. If a long barometer tube is placed in a dish of ethanol, with the top end open to the atmosphere, to what height in the tube will the ethanol rise? Atmospheric pressure = 1.01 * 105 Pa g = 9.81 m/s2 Density of water = 1000 kg/m3)

A diver in a lake rises from deep, where the gauge pressure reads , to the surface, where the gauge pressure reads . Upon reaching the surface, by what percent has the pressure experienced by the diver changed?

Two coins lie at the bottom of a pool with an uneven bottom. One is under of the water, and the second is under of water. What is the approximate ratio of total pressure experienced by the first coin to total pressure experienced by the second?

Atmospheric pressure always equals on earth, assuming normal elevation. Gauge pressure always equals , where is the density of the medium, is gravity, and is the distance to the surface.

A tank is filled to a depth of with water. Oil with a density of is added to the tank until the final depth is . What is the reading of the pressure gauge at the bottom of the tank? Ignore the contribution of atmospheric pressure.

Ethanol has a specific gravity of 0.780. If a long barometer tube is placed in a dish of ethanol, with the top end open to the atmosphere, to what height in the tube will the ethanol rise? Atmospheric pressure = 1.01 * 105 Pa g = 9.81 m/s2 Density of water = 1000 kg/m3)

A diver in a lake rises from deep, where the gauge pressure reads , to the surface, where the gauge pressure reads . Upon reaching the surface, by what percent has the pressure experienced by the diver changed?

Two coins lie at the bottom of a pool with an uneven bottom. One is under of the water, and the second is under of water. What is the approximate ratio of total pressure experienced by the first coin to total pressure experienced by the second?

Atmospheric pressure always equals on earth, assuming normal elevation. Gauge pressure always equals , where is the density of the medium, is gravity, and is the distance to the surface.

A tank is filled to a depth of with water. Oil with a density of is added to the tank until the final depth is . What is the reading of the pressure gauge at the bottom of the tank? Ignore the contribution of atmospheric pressure.

Ethanol has a specific gravity of 0.780. If a long barometer tube is placed in a dish of ethanol, with the top end open to the atmosphere, to what height in the tube will the ethanol rise? Atmospheric pressure = 1.01 * 105 Pa g = 9.81 m/s2 Density of water = 1000 kg/m3)

A diver in a lake rises from deep, where the gauge pressure reads , to the surface, where the gauge pressure reads . Upon reaching the surface, by what percent has the pressure experienced by the diver changed?

Two coins lie at the bottom of a pool with an uneven bottom. One is under of the water, and the second is under of water. What is the approximate ratio of total pressure experienced by the first coin to total pressure experienced by the second?

Atmospheric pressure always equals on earth, assuming normal elevation. Gauge pressure always equals , where is the density of the medium, is gravity, and is the distance to the surface.

A tank is filled to a depth of with water. Oil with a density of is added to the tank until the final depth is . What is the reading of the pressure gauge at the bottom of the tank? Ignore the contribution of atmospheric pressure.

Ethanol has a specific gravity of 0.780. If a long barometer tube is placed in a dish of ethanol, with the top end open to the atmosphere, to what height in the tube will the ethanol rise? Atmospheric pressure = 1.01 * 105 Pa g = 9.81 m/s2 Density of water = 1000 kg/m3)

A diver in a lake rises from deep, where the gauge pressure reads , to the surface, where the gauge pressure reads . Upon reaching the surface, by what percent has the pressure experienced by the diver changed?

Two coins lie at the bottom of a pool with an uneven bottom. One is under of the water, and the second is under of water. What is the approximate ratio of total pressure experienced by the first coin to total pressure experienced by the second?

Atmospheric pressure always equals on earth, assuming normal elevation. Gauge pressure always equals , where is the density of the medium, is gravity, and is the distance to the surface.

A tank is filled to a depth of with water. Oil with a density of is added to the tank until the final depth is . What is the reading of the pressure gauge at the bottom of the tank? Ignore the contribution of atmospheric pressure.

Ethanol has a specific gravity of 0.780. If a long barometer tube is placed in a dish of ethanol, with the top end open to the atmosphere, to what height in the tube will the ethanol rise?

Atmospheric pressure = 1.01 * 105 Pa

g = 9.81 m/s2

Density of water = 1000 kg/m3)

Two coins lie at the bottom of a pool with an uneven bottom. One is under $\small 5m$ of the water, and the second is under $\small 0.5m$ of water. What is the approximate ratio of total pressure experienced by the first coin to total pressure experienced by the second?

A tank is filled to a depth of $\small 3.27m$ with water. Oil with a density of $\small $800$\frac{kg}{m^3$}$$ is added to the tank until the final depth is $\small 5.0m$ . What is the reading of the pressure gauge at the bottom of the tank?

Ignore the contribution of atmospheric pressure.

Ethanol has a specific gravity of 0.780. If a long barometer tube is placed in a dish of ethanol, with the top end open to the atmosphere, to what height in the tube will the ethanol rise?

Atmospheric pressure = 1.01 * 105 Pa

g = 9.81 m/s2

Density of water = 1000 kg/m3)

Two coins lie at the bottom of a pool with an uneven bottom. One is under $\small 5m$ of the water, and the second is under $\small 0.5m$ of water. What is the approximate ratio of total pressure experienced by the first coin to total pressure experienced by the second?

A tank is filled to a depth of $\small 3.27m$ with water. Oil with a density of $\small $800$\frac{kg}{m^3$}$$ is added to the tank until the final depth is $\small 5.0m$ . What is the reading of the pressure gauge at the bottom of the tank?

Ignore the contribution of atmospheric pressure.

Ethanol has a specific gravity of 0.780. If a long barometer tube is placed in a dish of ethanol, with the top end open to the atmosphere, to what height in the tube will the ethanol rise?

Atmospheric pressure = 1.01 * 105 Pa

g = 9.81 m/s2

Density of water = 1000 kg/m3)

Two coins lie at the bottom of a pool with an uneven bottom. One is under $\small 5m$ of the water, and the second is under $\small 0.5m$ of water. What is the approximate ratio of total pressure experienced by the first coin to total pressure experienced by the second?

A tank is filled to a depth of $\small 3.27m$ with water. Oil with a density of $\small $800$\frac{kg}{m^3$}$$ is added to the tank until the final depth is $\small 5.0m$ . What is the reading of the pressure gauge at the bottom of the tank?

Ignore the contribution of atmospheric pressure.

Ethanol has a specific gravity of 0.780. If a long barometer tube is placed in a dish of ethanol, with the top end open to the atmosphere, to what height in the tube will the ethanol rise?

Atmospheric pressure = 1.01 * 105 Pa

g = 9.81 m/s2

Density of water = 1000 kg/m3)

Two coins lie at the bottom of a pool with an uneven bottom. One is under $\small 5m$ of the water, and the second is under $\small 0.5m$ of water. What is the approximate ratio of total pressure experienced by the first coin to total pressure experienced by the second?

A tank is filled to a depth of $\small 3.27m$ with water. Oil with a density of $\small $800$\frac{kg}{m^3$}$$ is added to the tank until the final depth is $\small 5.0m$ . What is the reading of the pressure gauge at the bottom of the tank?

Ignore the contribution of atmospheric pressure.

Ethanol has a specific gravity of 0.780. If a long barometer tube is placed in a dish of ethanol, with the top end open to the atmosphere, to what height in the tube will the ethanol rise?

Atmospheric pressure = 1.01 * 105 Pa

g = 9.81 m/s2

Density of water = 1000 kg/m3)

Two coins lie at the bottom of a pool with an uneven bottom. One is under $\small 5m$ of the water, and the second is under $\small 0.5m$ of water. What is the approximate ratio of total pressure experienced by the first coin to total pressure experienced by the second?

A tank is filled to a depth of $\small 3.27m$ with water. Oil with a density of $\small $800$\frac{kg}{m^3$}$$ is added to the tank until the final depth is $\small 5.0m$ . What is the reading of the pressure gauge at the bottom of the tank?

Ignore the contribution of atmospheric pressure.