A rectangular barge, 5.2 m long and 2.4 m wide, floats in fresh water. Suppose that a 410-kg crate of auto parts is loaded onto the barge. Part A) How much deeper does the barge float?
2 answers:
<h2>Answer:</h2><h2>The depth of barge float=
3 cm </h2><h2>
Explanation: </h2>
Length of rectangular barge=5.2 m
Width of rectangular barge=2.4m
Mass of crate=410 kg
Let h be the height of barge float
Volume of barge float=
Density of water=
Weight of water displaced by barge=Buoyant force=-Weight of horse
1 m=100 cm
cm
Hence, the depth of barge float=3 cm
<h2 />
Answer:
Explanation:
length, l = 5.2 m
width, w = 2.4 m
density of water = 1000 kg/m³
Let the depth immersed in water is h.
So, According to the principle of flotation
Weight of the auto parts = Buoyant force acting on the barge
410 x g = 5.2 x 2.4 x h x 1000 x g
410 = 12480 h
h = 3.3 cm
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