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mafiozo [28]
4 years ago
9

Which is more dense, cold freshwater or warm seawater?

Physics
2 answers:
svp [43]4 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]4 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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Original length = 2.97 m

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Acceleration due to gravity (g) = 9.8 m/s²

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Now, the length is shortened by 1.0 m. So, the new length is 1 m less than the original length.

New length of the pendulum is, L_1=L-1

New time period of the pendulum is, T_1=2.81\ s

We know that, the time period of a simple pendulum of length 'L' is given as:

T=2\pi\sqrt{\frac{L}{g}}-------------- (1)

So, for the new length, the time period is given as:

T_1=2\pi\sqrt{\frac{L_1}{g}}------------ (2)

Squaring both the equations and then dividing them, we get:

\dfrac{T^2}{T_1^2}=\dfrac{(2\pi)^2\frac{L}{g}}{(2\pi)^2\frac{L_1}{g}}\\\\\\\dfrac{T^2}{T_1^2}=\dfrac{L}{L_1}\\\\\\L=\dfrac{T^2}{T_1^2}\times L_1

Now, plug in the given values and calculate 'L'. This gives,

L=\frac{3.45^2}{2.81^2}\times (L-1)\\\\L=1.507L-1.507\\\\L-1.507L=-1.507\\\\-0.507L=-1.507\\\\L=\frac{-1.507}{-0.507}=2.97\ m

Therefore, the original length of the simple pendulum is 2.97 m

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Ciara is swinging a 0.015 kg ball tied to a string around her head in a flat, horizontal circle. The radius of the circle is 0.5
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Answer:

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