Question

16000 Pa, 19200

8. A rectangular boat is 4.0 m wide, 8.0 m long, and 3.0 m deep. (a) How much water will it
displace if the top stays 1.0 m above the water? (b) What load will the boat contain under
these conditions if the empty boat weighs 8.60× 104 N in dry dock?
(Weight density of water = 9800 N/m³)
(64.0 m³, 5.41x105 N)

Answer 1

The water displaced by a boat 4.0 m wide, 8.0 m long, 3.0 m deep, with 1.0 m above water, is 64.0 m³. The load the boat can carry, given that it weighs 8.60×10´ N in dry dock and with a water weight density of 9800 N/m³, is 5.41×10µ N.

Step-by-step explanation:

The problem described pertains to the concept of buoyant force as defined by Archimedes' principle, which states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid that the object displaces. The student's question involves calculating the water displaced by a boat and the load it can carry.

Calculating Water Displacement and Load

(a) To find the water displaced by the boat, we can calculate the submerged volume of the boat: 4.0 m (width) × 8.0 m (length) × 2.0 m (submerged depth) = 64.0 m³ of water.

(b) To find the load the boat can contain, we first need to determine the weight of the water displaced using the weight density of water (9800 N/m³). The weight of the water displaced is therefore 64.0 m³ × 9800 N/m³ = 6.272×10µ N. The maximum load the boat can carry without sinking is this value minus the weight of the empty boat (8.60×10´ N), so the boat can carry an additional load of 6.272×10µ N - 8.60×10´ N = 5.41×10µ N.

Answer 2

The rectangular boat will displace 64.0 m³ of water if the top stays 1.0 m above the water. The boat will contain a load of 5.41 x 10^5 N under these conditions if the empty boat weighs 8.60 x 10^4 N in dry dock.

Step-by-step explanation:

(a) The volume of the rectangular boat can be determined by multiplying its width, length, and depth: 4.0 m x 8.0 m x 3.0 m = 96.0 m³. To calculate how much water the boat will displace, we need to subtract the volume of the part of the boat above the water, which is 1.0 m x 4.0 m x 8.0 m = 32.0 m³. Therefore, the boat will displace 64.0 m³ of water.

(b) To find the load the boat will contain, we first need to determine its weight when it is empty. The weight of the boat in dry dock is given as 8.60 x 10^4 N. Since weight is the product of mass and gravity, we can rearrange this equation to find the mass: weight = mass x gravity. The weight density of water is given as 9800 N/m³, so the mass of the displaced water is mass_water = 64.0 m³ x 9800 N/m³ = 6.27 x 10^5 N. Adding the weight of the empty boat to the weight of the water gives us the total load: total_load = weight_empty_boat + weight_water = 8.60 x 10^4 N + 6.27 x 10^5 N = 5.41 x 10^5 N.

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