Question

One liter of water weighs about 10 n. suppose a 3-liter container is filled with 266N of mercury and then lowered into a container of water. Assuming the container is completely submerged, what is the buoyant force acting on it?

how do you get the answer to this problem?

Answer 1

The buoyant force acting on the 3-liter container submerged in water is 29.4 N.

Step-by-step explanation:

According to Archimedes' principle, the weight of water displaced is equal to the buoyant force acting on an object submerged in water.

In this case, the weight of the mercury-filled container is 266N. When it is submerged in water, the buoyant force acting on it will be equal to the weight of the water displaced.

Since the container has a volume of 3 liters, which is equal to 3 kilograms, and the density of water is approximately 1 kilogram per liter, the buoyant force can be calculated as follows:

Buoyant force = density of water x volume of water displaced x acceleration due to gravity

Buoyant force = 1 kg/L x 3 L x 9.8 m/s^2 = 29.4 N

Therefore, the buoyant force acting on the container is 29.4 N.

Answer 2

30 N

Step-by-step explanation:

The buoyant force acting on an object is an upward force exerted by the fluid in which the object is immersed.

The magnitude of the buoyant force is equal to the weight of the fluid displaced by the object, mathematically:

`B=\rho Vg`

where

`\rho` is the density of the fluid

V is the volume of fluid displaced

g is the acceleration due to gravity

The formula can be written also as

`B=mg`

where m is the mass of fluid displaced.

In this problem, the 3-Liter container filled with mercury is completely submerged in the water: this means that the volume of water displaced is

V = 3 L

We also know that 1 L of water weights about 10 N, so the weight of 3 L of water is

`mg=3\cdot 10 = 30 N`

And therefore, this is the buoyant force acting on the container: 30 N.