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mafiozo [28]
3 years ago
9

Which is more dense, cold freshwater or warm seawater?

Physics
2 answers:
svp [43]3 years ago
4 0
Cold freshwater<span> is </span>denser<span> than </span>warm seawater<span>, because of the salinity and temperature variations. Cold water would have less salt since the solubility of the salt is lower as compared to warm water. Hope this answers the question. Have a nice day.</span>
MAVERICK [17]3 years ago
4 0

Answer:

  • There is always the difference of energy that each atoms has inside its electronic configuration and this factor of the molecules or atoms provides the basic structure or arrangement of atoms inside the given space.As, the difference inn temperature,T or the total energy possessed by these number of molecules are the deciding factor inside the given space or region.
  • While, the atoms or molecules of the cold freshwater are said to be low on calcium ions and that they posses the energy in more compact form inside the inter molecular bonds, leaving the medium with less energy  in order to raise its temperature,T. So, the cold water has less temperature as compared to the warm seawater which posses more unsystematic-ally arranged atoms in the whole configuration of the medium.
  • And that is the reason that cold water are more dense or has more arranged or we can say that compact form of molecular arrangements inside there body, as compared to the warm water having less or proper arrangements of the molecules or atoms.

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mrs_skeptik [129]
Derived Units Table: The Table Shows the List of Derived Units
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5 0
2 years ago
A point charge with a charge q1 = 2.30 μC is held stationary at the origin. A second point charge with a charge q2 = -5.00 μC mo
Alla [95]

Answer:

W = 2.74 J

Explanation:

The work done by the charge on the origin to the moving charge is equal to the difference in the potential energy of the charges.

This is the electrostatic equivalent of the work-energy theorem.

W = \Delta U = U_2 - U_1

where the potential energy is defined as follows

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}

Let's first calculate the distance 'r' for both positions.

r_1 = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2} = \sqrt{(0.170 - 0)^2 + (0 - 0)^2} = 0.170~m\\r_2 = \sqrt{(x_2 - x_0)^2 + (y_2 - y_0)^2} = \sqrt{(0.250 - 0)^2 + (0.250 - 0)^2} = 0.353~m

Now, we can calculate the potential energies for both positions.

U_1 = \frac{kq_1q_2}{r_1^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.170)^2} = -3.57~J\\U_2 = \frac{kq_1q_2}{r_2^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.3530)^2} = -0.829~J

Finally, the total work done on the moving particle can be calculated.

W = U_2 - U_1 = (-0.829) - (-3.57) = 2.74~J

4 0
3 years ago
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mestny [16]

This is what I got:

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T1 - ma = T2

100 - (3)(2) = T2

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