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allochka39001 [22]
3 years ago
13

What % of an object’s mass is above the water line if the object’s density is 0.82g/ml?

Physics
1 answer:
Sladkaya [172]3 years ago
4 0

To develop this problem we will apply the Archimedes model. As well as the definitions of Weight based on mass and acceleration. The first in turn will be considered under the relationship of Density and Volume. From the values given we have to:

\rho_w =1 g/mL \rightarrow \text{Water Density}

\rho_o = 0.82g/mL \rightarrow \text{Object density}

Since it is in equilibrium, the weight of the object will have a reaction from the water, which will cause the sum of forces between the two objects to be zero, therefore

\sum F= 0

F_w-F_o = 0

F_w = F_o

m_w g = m_og

\rho_w V_w g = \rho_o V_o g

The value of gravity is canceled because it is a constant

\frac{V_w}{V_o} = \frac{\rho_o}{\rho_w}

\frac{V_w}{V_o} = \frac{0.82}{1}

\frac{V_w}{V_o} = 0.82

The portion of the object that is submerged corresponds to 82%, while the portion that is visible, above the water level will be 18%

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Answer:

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Given that,

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Using equation of motion

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Put the value into the formula

t = \dfrac{80-0}{3}

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The time is 26.67 sec.

(b). We need to calculate the distance traveled during this period

Using equation of motion

s = ut+\dfrac{1}{2}at^2

s = 0+\dfrac{1}{2}\times3\times(26.67)^2

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The distance traveled during this period is 1066.9 m.

Hence, This is the required solution.

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A cannon placed at the origin fires a projectile with velocity ~v0, which passes during its trajectory through two points both a
FinnZ [79.3K]

Answer:

The equations for the y position of the trajectory is:

y=sin\phi v_0t-\frac{1}{2}gt^2

The range is of the trajectory is reached at y = 0:

sin\phi v_0t-\frac{1}{2}gt^2 = 0

Solving for time t:

t=\frac{2sin\phi v_0}{g}

The x position of the trajectory is given by:

x=cos\phi v_0t

Combining the equations we get the function:

x=\frac{2sin\phi cos\phi v_0^2}{g}=sin(2\phi)\frac{v_0^2}{g}

Taking the derivative and setting it to zero:

\frac{dx}{d\phi}=2cos(2\phi)\frac{v_0^2}{g}= 0

Find the maximum angle:

\phi=45

Using this solution to find the trajectory in terms of x and y and setting it equal to height h:

y=\frac{sin45 v_0}{cos45v_0}x-\frac{g}{2cos^245v_0^2}x^2=x-\frac{g}{v_0^2}x^2=h

Normalize:

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Use quadratic formula:

x_{1/2}=\frac{v_0^2}{2g}+/-v_0\sqrt{\frac{v_0^2-4hg}{4g^2}}

The distance is the difference between the two points:

d=|x_1-x_2|

d=\frac{v_0}{g} \sqrt{v_0^2-4hg}

5 0
3 years ago
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