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inysia [295]
3 years ago
8

A boat is 4.0meters long and 2.0meters wide. Its water line is 1.0meter from the bottom of the boat; it is submerged to this wat

er line. What is the mass of the boat? A. 8000kgB. 600kgC. 130kgD. 800kgE. 200kg
Physics
1 answer:
Mademuasel [1]3 years ago
3 0

Answer:

option A

Explanation:

given,

Length of boat = 4 m

width of boat = 2 m

height up to which water is displaced = 1 m

Volume of water displaced by the boat

 V =  L x B x H

 V = 4 x 2 x 1

 V = 8 m³

buoyant force acting on the boat

F_b = ρ g V

m g = ρ g V

 m = ρ V

ρ is the density of water which is equal to 1000 Kg/m³

 m = ρ V

 m =1000 x 8

 m = 8000 Kg

hence, mass of the boat is equal to 8000 Kg

so, the correct answer is option A

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h=\dfrac{E}{mg}\\\\h=\dfrac{520}{2.2\times 9.8}\\\\h=24.11\ m

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Now we’ll use the component method to add two vectors. We will use this technique extensively when we begin to consider how forc
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The magnitude of the vector sum of A and B is 65.8 cm and its direction 61.6°

Explanation:

Since vector A has magnitude 50 cm and a direction of 30, its x - component is A' = 50cos30 = 43.3 cm and its y - component is A" = 50sin30 = 25.

Also, Since vector B has magnitude 35 cm and a direction of 110, its x - component is A' = 35cos110 = -11.97 cm and its y - component is A" = 35sin110 = 32.89 cm.

So, the vector sum R = A + B

The x-component of the vector sum is R' = A'+ B' = 43.3 cm + (-11.97 cm) = 43.3 cm - 11.97 cm = 31.33 cm

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So, the magnitude of R = √(R'² + R"²)

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= √(4,332.821 cm²)

= 65.82 cm

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The direction of R is Ф = tan⁻¹(R"/R')

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= 61.58°

≅ 61.6°

So, the magnitude of the vector sum of A and B is 65.8 cm and its direction 61.6°

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3 years ago
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