1. Suppose a tank filled with water has a liquid column with a height of 10 meters. If the area is 2 square meters (m²), what's

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2 months ago

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1. Suppose a tank filled with water has a liquid column with a height of 10 meters. If the area is 2 square meters (m²), what's

the force of gravity acting on the column of water?
2. If a total force exerted by water in a container with a bottom area of 3 square meters is 900 newtons, what is the water pressure at the bottom of the container?
3. A tank with a flat bottom is filled with water to a height of 7.5 meters. What's the pressure at any point at the bottom of the tank? (You can ignore atmospheric pressure in your calculations.)
4. In a tank full of water, the pressure on a surface 2 meters below the water level is 1.5 kPa. What's the pressure on a surface 6 meters below the water level?
5. A piston above a liquid in a closed container has an area of 0.75 m², and the piston carries a load of 200 kg. What will be the external pressure on the upper surface of the liquid?

Physics

2 answers:

Ostrovityanka [3.2K] · Answer 1 · 2026-03-21T16:43:47+03:00

Answer: 1. F = Ahdg

F = 2 m2 × 10 m × 1,000 kg/m3 × 9.8 m/s2

F = 20 × 1,000 × 9.8

F = 20,000 × 9.8

F = 196,000 N

2. P=FA

P

=

F

A

P=900N3m2

P

=

900

N

3

m

2

P = 300 Pa

This could also be expressed as 0.300 kPa. To shift from pascals to kilopascals, just divide by 1,000.

3. P = hdg

P = 7.5

P = 7.5 m × 1,000 kg/m2 × 9.8 m/s2

P = 73,500 pascals

This can also be represented as 73.5 kPa. Converting pascals to kilopascals requires dividing by 1,000.

4. The water column height at 6 meters beneath the water surface is three times greater. To find the pressure at this depth, multiply the pressure at 2 meters by 3.

P = 3 × 1.5 kPa

P = 4.5 kPa

5. P=mgA

P

=

m

g

A

P=200kg×9.8m/s20.75m/s2

P

=

200

k

g

×

9.8

m

/

s

2

0.75

m

/

s

2

P=1,9600.75

P

=

1

,

960

0.75

P = 2,613 Pa or 2.613 kPa

Explanation: pen foster

ValentinkaMS [3.4K] · Answer 2 · 2026-03-21T16:44:26+03:00

Answers:

1. To begin with, we need to clarify that Pressure $P$ is defined as the Force applied $F$ divided by the area $A$ . This can be mathematically formulated as follows:

$P=\frac{F}{A}$ (1)

The unit for Pressure is Pascal (Pa), which corresponds to $\frac{kg}{ms^{2}}$ and is also equivalent to $\frac{N}{m^{2} }$

Another formulation for Pressure considers the density $d$ of the liquid, its height $h$ within the container, and the force due to gravity $g$ :

$P=d*h*g$ (2)

In this scenario, the liquid in question is water, with a known density of roughly:

$d=1000kg/m^{3}$

Thus, we need to input the values into equation (2) to find the pressure (make sure to check the units):

$P=1000\frac{kg}{m^{3}}*10m*9.8\frac{m}{s^{2}}$

$P=98000Pa$

Next, substitute this value into equation (1) to isolate $F$ :

$F=P*A$

In conclusion:

$F=196000N$

2. For this problem, we will apply equation (1) to determine the Pressure. We already have the area $A$ and the force applied by the water in the container $F$ :

$P=\frac{F}{A}=\frac{900N}{3m^{2}}$

$P=300Pa$

3. Here, equation (2) is ideal for calculating the hydrostatic pressure at any point located at the bottom of the tank (be cautious with your units):

$P=d*h*g$

$P=1000\frac{kg}{m^{3}}*7.5m*9.8\frac{m}{s^{2}}$

$P=73500Pa$

4. In this situation, it's critical to note that in fluids (specifically water here), higher fluid levels result in reduced pressure. Therefore, if $P_{1}$ and $P_{2}$ denote the pressures at heights $h_{1}$ and $h_{2}$ , respectively, and recognizing that the density of water and gravitational force remain constant, we can employ the following equation for our solution: