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monitta
3 years ago
15

Does warm water hold more salt then cold water?

Physics
2 answers:
maksim [4K]3 years ago
8 0

yes. ♛┈⛧┈┈•༶༶•┈┈⛧┈♛♛┈⛧┈┈•༶༶•┈┈⛧┈♛

Vinil7 [7]3 years ago
3 0

Answer: Yes

Explanation: Because the water molecules expand when exposed the the warmth of the sun witch means that warm water can carry more salt.

Hope this helps!

Have a good rest of your day!

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I believe the answer is the second option.
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A 5.00 kg rock whose density is 4300 kg/m3 is suspended by a string such that half of the rock's volume is under water. You may
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Answer:

The tension on the string is  T  =  43.302 \ N

Explanation:

From the question we are told that

    The mass of the rock is m_r = 5.00 \ kg =  5000 \ g

       The density of the rock is \rho  =  4300 \ kg/m^3 =  4.3 g/dm^3

       

Generally the volume of the rock is mathematically evaluated as

          V    =  \frac{m_r}{\rho}

substituting values

        V    =  \frac{5000}{4.3}

       V    =  1162.7 \  dm^3

The volume of the rock immersed in water is

      V_w = \frac{V}{2}  

substituting values

     V_w = \frac{1162.7 }{2}

     V_w = 581.4 \ dm^3

mass of water been displaced by the this volume is

     m_w  = V_w     According to Archimedes principle

=>   m_w =  581.4 \ g

     m_w =  0.5814 \ kg

The weight of the water displace is  

      W _w =  m_w  * g

      W _w =  0.5814  * 9.8

      W _w = 5.698 \ N

The actual weight of the rock is  

      W_r  =  m_r * g

     W_r  =  5.0 *  9.8

     W_r  =  49.0 \ N

The tension on the string is

       T  = W_r - W_w

substituting values

       T  = 49.0 -  5.698

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2 years ago
A lady bug is sitting on the bottom of a can while you twirl it overhead on a string that is 65.0
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The linear speed of the ladybug is 4.1 m/s

Explanation:

First of all, we need to find the angular speed of the lady bug. This is given by:

\omega=\frac{2\pi}{T}

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The period of revolution is the time taken by the ladybug to complete one revolution: in this case, since it does 1 revolution every second, the period is 1 second:

T = 1 s

Therefore, the angular speed is

\omega=\frac{2\pi}{1 s}=6.28 rad/s

Now we can find the linear speed of the ladybug, which is given by

v=\omega r

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Substituting, we find

v=(6.28)(0.65)=4.1 m/s

Learn more about angular speed:

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brainly.com/question/2506028

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