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Natasha_Volkova [10]

4 years ago

15

According to Archimedes' principle, the mass of a floating object equals the mass of the fluid displaced by the object. A 150-lb

m swimmer is floating in a nearby pool; 95% of his or her body's volume is in the water while 5% of his or her body's volume is above water. Determine the density of the swimmer's body. The density of water is 0.036lbm/in^3.

1 answer:

Archy [21]4 years ago

3 0

Answer:

Density of the swimmer = $0.0342\ lbm/in^3$ .

Explanation:

Assuming,

$\rho$ = density of the swimmer.

$\rho_w$ = density of the water.

$m$ = mass of the swimmer.

$m_w$ = mass of the water displaced by the swimmer.

$V_w$ = volume of the displaced water.

$V$ = volume of the swimmer.

Given:

$m=150\ lbm.$
$\rho_w = 0.036\ lbm/in^3.$

The density of an object is defined as the mass of the object per unit volume.

Therefore,

$\rho =\dfrac{m}{V}\ \Rightarrow m = \rho V\ \ .........\ (1).$

Since only 95% of the body of the swimmer is inside the water, therefore,

$V_w = 95\%\ \text{of}\ V=\dfrac{95}{100}\times V = 0.95V.$

According to Archimedes' principle,

$m=m_w\\$

Using (1),

$\rho V=\rho_w V_w\\\rho V = 0.036\ lbm/in^3\times 0.95 V\\\rho=0.036\times 0.95\ lbm/in^3=0.0342\ lbm/in^3.$

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The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

Explanation:

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Suppose, the following information would students need to test the hypothesis,

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The work done by a force is given by,

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We know, the net force is the sum of forces.

So, $\sum{F}=ma$

According to question,

We have two forces F(x) and F(f)

So, the sum of these forces are

$F(x)+(-F(f))=ma$

Here, frictional force is negative because F(f) acts against the F(x)

Now put the value in equation (I)

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