I am new at physics and having a very hard time. Could you please help. Thank you.

Show that the water pressure at the bottom of the 50-m-high water tower in Figure 5.3 is 490,000 N/m2, or is approximately 500 kPa.

Sure! To solve this problem, we can use the formula for pressure:

Pressure = Density × Gravity × Height

First, let's determine the density of water. The density of water is approximately 1000 kg/m³.

Next, we need to identify the value of gravity. On Earth, the acceleration due to gravity is approximately 9.8 m/s².

Lastly, we need to find the height of the water tower, which is given as 50 meters.

Plugging these values into the formula, we get:

Pressure = 1000 kg/m³ × 9.8 m/s² × 50 m

Calculating this equation, we find:

Pressure = 490,000 N/m²

Therefore, the water pressure at the bottom of the 50-meter-high water tower is 490,000 N/m², or approximately 500 kPa.

Of course, I'd be happy to help you with this physics problem!

To determine the water pressure at the bottom of the 50-meter-high water tower, we can use the concept of hydrostatic pressure. Hydrostatic pressure is the pressure exerted by a fluid at rest due to the weight of the fluid above it. In this case, the fluid is water and the height of the water tower provides the basis for calculation.

Here's how you can calculate the water pressure at the bottom of the water tower:

1. Understand the concept: The relationship between pressure, height (depth), and the density of a fluid is given by the equation: P = ρgh, where P is the pressure, ρ (rho) is the density of the fluid, g is the acceleration due to gravity, and h is the height (depth) of the fluid column.

2. Gather the required information: From the problem statement, we are given that the height of the water tower, h, is 50 meters. We also need the density of water, which is approximately 1000 kg/m³, and the gravitational acceleration, which is approximately 9.8 m/s².

3. Calculate the pressure: Using the equation P = ρgh, we can substitute the known values:

P = (1000 kg/m³) × (9.8 m/s²) × (50 m)
= 490,000 N/m²

Therefore, the water pressure at the bottom of the 50-meter-high water tower is 490,000 N/m² or 490,000 pascals (Pa). We can also approximate this value as approximately 500 kilopascals (kPa).

If you have any further questions regarding this problem or any other physics concepts, feel free to ask!

One way:

How much does a column of water one meter square at the bottom and 50 m high weigh?

Density of water = 1000 Kg /m^3
Volume of column = 50*1 = 50 m^3
so
mass of the column = 50,000 Kg
weight = m g = 9.81 * 50,000 = 490,500 Newtons spread over one m^2 so
490,000 N/m^2

another way
pressure = rho g h
= 1000 * 9.81 * 50 = 490,500 N/m^2 again
since a Pascal is a Newton/meter^2 we have already answered the second part, about 500 * 10^3 Pascals