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Serhud [2]
3 years ago
11

An object weighing 2.7n while in air and 1.2n when completely immersed in water.find the relative density of the object

Physics
1 answer:
Elden [556K]3 years ago
8 0

Answer:

1.8 = relative density (there are no units for relative density)

Explanation:

It displaces water equal to it's volume and gets buoyancy equal to that amount of water

2.7 - 1.2  = 1.5  N  of buoyancy

density of water = 1 gm /cc

 1.5 N = m (9.81)

    m of water displaced = .1529 kg

             152.9 cc of water will produce this buoyancy....this is the volume of the object

   

find mass of object     2.7 = m (9.81)     shows m = .2752 kg = 272.5 gm

density = mass/ volume = 272.5 / 152.9 = 1.8 gm / cc

Relative to water (which is 1 gm / cc) the relative density is 1.8

  ====> it is 1.8 times denser than water and will sink when in water....

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4 years ago
What is the energy difference between parallel and antiparallel alignment of the z component of an electron's spin magnetic dipo
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Answer:

\Delta U= 4.8204\times10^{-24} J

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here θ_2 and θ_1 are 180 and 0° respectively.

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put this value we get

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Consider the following distribution of objects: a 5.00-kg object with its center of gravity at (0, 0) m, a 3.00-kg object at (0,
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Answer:

(-1.5,-1.5)m

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we know that:

X_{cm} = \frac{m_1x_1+m_2x_2....m_nX_n}{m_1+m_2...m_n}

where X_{cm} is the location of the center of gravity in the axis x, m_i is the mass of the object i and x_i the first coordinate of center of gravity of object i.

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0 = \frac{(5kg)(0)+(3kg)(0)+(4kg)(3)+(8kg)x_4}{5kg+3kg+4kg+8kg}

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At the same way:

Y_{cm} = \frac{m_1y_1+m_2y_2....m_ny_n}{m_1+m_2...m_n}

where Y_{cm} is the location of the center of gravity in the axis y, m_i is the mass of the object i and y_i the second coordinate of center of gravity of object i. replacing values we get:

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solving for y_4:

y_4 = -1.5m

It means that the object of mass 8kg have to be placed in the  

coordinates (-1.5,-1.5) m.

8 0
3 years ago
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