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BlackZzzverrR [31]
3 years ago
14

a 55.0 kg person having a denity of 1010 kg/m^3 is underwater. What is the net force acting on the person?

Physics
1 answer:
bija089 [108]3 years ago
6 0

Answer:

The net force acting on the person is 5.34 N acting upwards

Explanation:

The information given are;

Mass of the person = 55.0 kg

Density of the person = 1010 kg/m³

Body volume of the person = Mass/Density = 55/1010 = 0.0545 m³

Weight of the person = Mass × Gravity = 55 × 9.81 = 539.55 N↓

Where the density of water = 1000 kg/m³, we have;

Upthrust = weight of the water displaced = Mass of the water displaced × Gravity

Mass of the water displaced = Density of water × Volume of the water

Mass of the water displaced = Density of water × Body volume of the person

                                                = 1000 × 0.0545 = 54.5 kg

The Upthrust = 54.5 × 9.81 = 534.21 N↑

Given that the net force, F_{Net} acting is the sum of the mass of forces acting  on the person which are the weight acting downward and the upthrust acting upwards

We have;

F(Net) = 534.21↑ - 539.55↓ = 5.34 N↑

The net force acting on the person = 5.34 N upwards.

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Answer:

θ = 30⁰  

Explanation:

We have been given two wave displacement equations,

y_{1} =2sin\theta +1     ........(1)

y_{2}=3sin\theta +0.5     ........(2)

The waves will have same displacements when y_{1}=y_{2}=y (say)

Therefore, operating (2) - (1), we get,

sin\theta - 0.5=0

or, \theta = sin^{-1}(0.5)=30^{o} (since, sin 30⁰ = 0.5)

We can check the answer by putting \theta=30^{o} in equations (1) and (2),

y_{1}=2sin30^{o}+1=2\times0.5 + 1= 2

y_{2}=3sin30^{o}+0.5=3\times0.5+0.5=1.5+0.5=2

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Dependent variable:

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