Question

.A cargo barge is loaded in a saltwater harbor for a trip up a freshwater river.

If the rectangular barge is 3.50 m by 24.0 m and sits 0.600 m deep in the harbor, how deep will it sit in the river?
h=________m

Answer 1

The cargo barge will sit 0.615 m deep in the river.

Step-by-step explanation:

To determine how deep the cargo barge will sit in the river, we need to understand the concept of buoyancy. Buoyancy is the upward force exerted on an object submerged in a fluid, such as water. It is equal to the weight of the fluid displaced by the object.

In this case, the weight of the barge remains the same regardless of whether it is in saltwater or freshwater. However, the density of saltwater is higher than freshwater, which means that the barge will displace less water in the river compared to the harbor.

To calculate the depth at which the barge will sit in the river, we can use the principle of Archimedes' principle. The volume of the barge can be calculated by multiplying its width, length, and depth. As the weight of the barge remains constant, the volume displaced will also remain constant. Therefore, we can use the formula:

Volume in Harbor = Volume in River

(Width in Harbor)(Length in Harbor)(Depth in Harbor) = (Width in River)(Length in River)(Depth in River)

Given that the dimensions of the barge are 3.50 m by 24.0 m and it sits 0.600 m deep in the harbor, we can substitute these values into the equation:

(3.50 m)(24.0 m)(0.600 m) = (3.50 m)(24.0 m)(Depth in River)

Solving for the depth in the river:

Depth in River = (3.50 m)(24.0 m)(0.600 m) / (3.50 m)(24.0 m)

Depth in River = 0.615 m