A very powerful vacuum cleaner has a hose 2.86 cm in diameter. With the end of the hose placed perpendicularly on the flat face of a brick, what is the weight of the heaviest brick that the cleaner can lift? (b) What If? An octopus uses one sucker of diameter 2.86 cm on each of the two shells of a clam in an attempt to pull the shells apart. Find the greatest force the octopus can exert on a clamshell in salt water 32.3 m deep.
To answer both questions, we need to understand the concepts of pressure and buoyancy.
1. Weight of the Heaviest Brick the Vacuum Cleaner Can Lift:
To calculate the weight of the heaviest brick the vacuum cleaner can lift, we need to determine if the suction force provided by the cleaner's hose is enough to overcome the weight of the brick. This can be done by comparing the pressure difference between the inside and the outside of the hose.
Assuming the vacuum cleaner creates a perfect vacuum inside the hose, the pressure difference is given by the equation:
ΔP = P_outside - P_inside
The pressure inside the hose (P_inside) is zero, as it's a vacuum. The pressure outside the hose (P_outside) is atmospheric pressure, which is approximately 1 atm.
Next, we need to calculate the force exerted by the pressure difference using the equation:
Force = Pressure * Area
The area in this case is the cross-sectional area of the hose, given by the formula:
Area = π * (radius)^2
First, let's calculate the radius of the hose:
radius = diameter / 2 = 2.86 cm / 2 = 1.43 cm = 0.0143 m
Now, substitute the values into the equation to find the area:
Area = π * (0.0143 m)^2
Once we have the area, we can calculate the force:
Force = ΔP * Area = (1 atm) * Area
Finally, to find the weight of the brick the vacuum cleaner can lift, we equate it with the force:
Weight_of_brick = Force
After calculating the force, you'll have the weight of the heaviest brick the vacuum cleaner can lift.
2. Greatest Force the Octopus Can Exert on a Clamshell in Salt Water:
To determine the greatest force the octopus can exert on a clamshell, we need to consider the pressure exerted by the octopus' suckers and the pressure difference caused by the depth of the water.
The pressure exerted by the octopus' sucker is given by:
Pressure = Force / Area
The area here refers to the cross-sectional area of the sucker, which can be calculated using the same formula as before:
Area = π * (radius)^2
Now, let's calculate the radius:
radius = 2.86 cm / 2 = 1.43 cm = 0.0143 m
Next, we need to account for the pressure difference caused by the depth of the water. The pressure difference due to depth is given by:
ΔP = Density of water * gravitational acceleration * depth
The density of water (ρ) and the gravitational acceleration (g) can be looked up or taken as constants:
ρ = 1000 kg/m³ (for seawater)
g = 9.8 m/s²
Substituting the values, we get:
ΔP = (1000 kg/m³) * (9.8 m/s²) * (32.3 m)
Now, we can calculate the force exerted by the octopus' suckers by multiplying the pressure with the area:
Force = Pressure * Area
Finally, you'll have the greatest force the octopus can exert on a clamshell.
Remember, always double-check your calculations and use appropriate units.