Water has a density of 1000kg/m^3. The depth of a swimming pool at the deep end is about 4m. A)What is the volume of a column of water 4m deep and 1m^3 in cross-sectional area? B)What is the mass of this column of water? C)What is the weight of this column of water in newtons? D)What is the excess pressure (above atmospheric pressure)exerted by this column of water on the bottom of the pool? E)How does this value compare to atmospheric pressure? I don't know where to begin, I am so confused by this question help!!!
A. V = 4m * 1m^2 = 4 m^3.
B. Mass = 4m^3 * 1000kg/m^3 = 4000 kg.
C. Wt. = 4000kg * 9.8N/kg = 39,200 N.
Don't worry, I'll help you step-by-step to solve each part of the question:
A) To find the volume of a column of water with a cross-sectional area of 1 m^2 and a depth of 4 m, you can use the formula:
Volume = Area x Depth
In this case, the area is 1 m^2 and the depth is 4 m. So, the volume of the column of water is:
Volume = 1 m^2 x 4 m = 4 m^3
B) To find the mass of the column of water, we can use the formula:
Mass = Density x Volume
The given density of water is 1000 kg/m^3 and the volume we found in part A is 4 m^3. So, the mass of the column of water is:
Mass = 1000 kg/m^3 x 4 m^3 = 4000 kg
C) To find the weight of the column of water, we can use the formula:
Weight = Mass x Gravity
The acceleration due to gravity is approximately 9.8 m/s^2. So, the weight of the column of water is:
Weight = 4000 kg x 9.8 m/s^2 = 39200 N
D) The excess pressure exerted by the column of water on the bottom of the pool can be calculated using the formula:
Excess Pressure = Density x Gravity x Depth
The given density, as mentioned earlier, is 1000 kg/m^3, the acceleration due to gravity is 9.8 m/s^2, and the depth is 4 m. Therefore, the excess pressure exerted on the bottom of the pool is:
Excess Pressure = 1000 kg/m^3 x 9.8 m/s^2 x 4 m = 39200 Pa
E) Atmospheric pressure is approximately 101325 Pa. To compare the excess pressure to atmospheric pressure, we can subtract atmospheric pressure from the excess pressure calculated in part D:
Difference = Excess Pressure - Atmospheric Pressure
Difference = 39200 Pa - 101325 Pa = -62125 Pa
The negative sign indicates that the exerted pressure is lower than atmospheric pressure.
I hope this helps you understand the steps involved in solving this question. If you have any more questions, feel free to ask!
No problem! Let's break down this question step by step and understand how to find the answers:
A) To find the volume of a column of water, we can use the formula: Volume = Area × Depth.
In this case, the area of the column is given as 1m^3, and the depth is 4m. So, the volume is:
Volume = 1m^3 × 4m = 4m^3.
B) To find the mass of the column of water, we can use the formula: Mass = Density × Volume.
The density of water is given as 1000kg/m^3, and the volume we found in part A is 4m^3. So, the mass is:
Mass = 1000kg/m^3 × 4m^3 = 4000kg.
C) To find the weight of the column of water, we can use the formula: Weight = Mass × Gravity.
The mass we found in part B is 4000kg, and the acceleration due to gravity is typically taken as 9.8m/s^2. So, the weight is:
Weight = 4000kg × 9.8m/s^2 = 39200N.
D) To find the excess pressure exerted by the column of water on the bottom of the pool, we can use the formula: Pressure = Density × Gravity × Depth.
The density of water is still 1000kg/m^3, the acceleration due to gravity is 9.8m/s^2, and the depth is 4m. So, the pressure is:
Pressure = 1000kg/m^3 × 9.8m/s^2 × 4m = 39200Pa (Pascals).
E) Atmospheric pressure is typically around 101325Pa. To compare the excess pressure to atmospheric pressure, we can simply subtract the atmospheric pressure from the excess pressure:
Excess pressure - Atmospheric pressure = 39200Pa - 101325Pa = -62125Pa.
The negative sign indicates that the excess pressure is lower than atmospheric pressure, which means the column of water is not exerting enough pressure to balance the atmospheric pressure.