How much force is needed to hold a nearly weightless but rigid 1-L carton beneath the surface of water?
it displaces 1 liter of water, which weighs 1kg*9.8N/kg, so force is 9.8N
To calculate the force needed to hold a nearly weightless but rigid 1-L carton beneath the surface of water, we can use Archimedes' principle.
1. Determine the weight of the water displaced: The force required to hold the carton underwater is equal to the weight of the water displaced by the carton.
2. Calculate the volume of water displaced: The volume of water displaced is equal to the volume of the carton, which is given as 1L (or 1,000 cm³).
3. Convert the volume to mass: Since the carton is nearly weightless, we can assume its mass is negligible. Therefore, the mass of the water displaced will be equal to the volume of water displaced.
4. Convert the mass to weight: Use the formula weight = mass x acceleration due to gravity. The acceleration due to gravity is approximately 9.8 m/s².
5. Calculate the force: Force is equal to weight. Thus, the force needed to hold the carton underwater is equal to the weight of the water displaced.
Therefore, to hold a nearly weightless but rigid 1-L carton beneath the surface of water, you would need a force equal to the weight of the water displaced, which can be calculated using the above steps.
To determine the force needed to hold a nearly weightless but rigid 1-L carton beneath the surface of water, you need to consider the buoyancy force and the weight of the carton.
The buoyancy force is the upward force exerted by a fluid on an immersed object. It is equal to the weight of the fluid displaced by the object, and it acts in the opposite direction to gravity.
Since the carton is nearly weightless, we can assume that its weight is negligible. Therefore, the force needed to hold the carton beneath the water's surface is equal to the buoyancy force.
To calculate the buoyancy force, you need to determine the weight of the fluid displaced by the carton. The weight of the fluid can be calculated using the formula:
Weight of the fluid = density of the fluid × volume of the fluid × acceleration due to gravity
In this case, the fluid is water, which has a density of approximately 1000 kg/m³. The volume of the fluid displaced is equal to the volume of the carton, which is 1 L or 0.001 m³. The acceleration due to gravity is approximately 9.8 m/s².
Now, let's calculate the weight of the fluid:
Weight of the fluid = 1000 kg/m³ × 0.001 m³ × 9.8 m/s²
Weight of the fluid = 9.8 N
Therefore, the force needed to hold the nearly weightless but rigid 1-L carton beneath the surface of water is approximately 9.8 Newtons.