Final answer:
The water force exerted on the bottom of a restaurant's lobster tank, which measures 0.75 m by 0.5 m by 0.7 m deep, is 2575.125 newtons.
Step-by-step explanation:
Calculating the Water Forces on the Bottom of a Restaurant's Lobster Tank
The question is asking us to calculate the force exerted by the water on the bottom of a lobster tank, given the dimensions and the density of water. To do this, we need to use the concept of pressure, which is defined as force per unit area. The pressure exerted by a static fluid in an open container depends on the density of the fluid (ρ), the acceleration due to gravity (g), and the depth of the fluid (h). The force on the bottom of the tank is found using the equation:
F = p • A
Where p is the pressure at the bottom of the tank, and A is the area of the bottom of the tank. Since the tank is open to the atmosphere, the pressure at the bottom due to the water column is:
p = ρ • g • h
The dimensions of the tank are 0.75 m by 0.5 m by 70 cm deep (which is 0.7 m). Assuming that the density of water (ρ) is 1000 kg/m³ and acceleration due to gravity (g) is approximately 9.81 m/s²:
p = 1000 kg/m³ • 9.81 m/s² • 0.7 m = 6867 N/m²
The area (A) of the tank's bottom is:
A = length • width = 0.75 m • 0.5 m = 0.375 m²
The water force (F) on the bottom of the tank is:
F = p • A = 6867 N/m² • 0.375 m² = 2575.125 N
So the force exerted by the water on the bottom of the lobster tank is 2575.125 newtons.