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Elanso [62]
3 years ago
5

Would the weight density of water be different on the moon than it is on earth?

Physics
1 answer:
-BARSIC- [3]3 years ago
6 0
No the weight would be changed due to the different gravity. The density won't change though. This is due to the density is related to the hydrogen bond between water
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PLEASE HELP! I NEED TO ANSWER THIS QUICKLY!
rodikova [14]

Answer:

I'm not a genius ok?

Explanation:

1. Radar communication, Analysis of the molecular and atomic structure, telephone communication

2. c

4 0
3 years ago
Which of the following has the least impact on the environment?
seraphim [82]
The answer should be A) solar power
8 0
4 years ago
Read 2 more answers
In general, if the temperature of a chemical reaction is increased, the reaction rateA. IncreasesB. decreasesC. remains the same
In-s [12.5K]
A. Increases

I would assume this to be the answer because heat is another form of energy. If there is more energy the molecules will become more active. This makes A the most logical answer.
6 0
3 years ago
Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its
ruslelena [56]

Answer:

a

The orbital speed is v= 2.6*10^{3} m/s

b

The escape velocity of the rocket is  v_e= 3.72 *10^3 m/s

Explanation:

Generally angular velocity is mathematically represented as

            w = \frac{2 \pi}{T}

Where T is the period which is given as 1.6 days = 1.6 *24 *60*60 = 138240 sec

       Substituting the value

         w = \frac{2 \pi}{138240}

             = 4.54*10^ {-5} rad /sec

At the point when the rocket is on a circular orbit  

   The gravitational force =  centripetal force and this can be mathematically represented as

              \frac{GMm}{r^2} = mr w^2

Where  G is the universal gravitational constant with a value  G = 6.67*10^{-11}

            M is the mass of the earth with a constant value of M = 5.98*10^{24}kg

            r is the distance between earth and circular orbit where the rocke is found

               Making r the subject

                     r = \sqrt[3]{\frac{GM}{w^2} }

                        = \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }

                        = 5.78 *10^7 m

The orbital speed is represented mathematically as

                   v=wr

Substituting value

                  v= (5.78*10^7)(4.54*10^{-5})

                     v= 2.6*10^{3} m/s    

The escape velocity is mathematically represented as

                            v_e = \sqrt{\frac{2GM}{r} }

Substituting values

                             = \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }

                             v_e= 3.72 *10^3 m/s

7 0
3 years ago
A block of wood of mass 24 kg floats on water. The volume of the block below the surface of the water and the density of the woo
Slav-nsk [51]

Answer:

0.024m^3

Explanation:

=======

Answer:

=======

Given:-

Mass of the block of wood = 24 kg

Volume of wood = 0.032 m^3

Density of water = 1000kg/m^3

Now,

Density of wood is given by,

\frac{m}{v} = \frac{24}{0.032} \\

\frac{m}{v} = 750 \: kg/m ^{3}

Therefore,

The density of wood is 750kg/m^3

By principle of floatation,

Mass \:of\: wood = Mass\: of\: liquid \:displaced

Therefore,

Mass of liquid displaced = 24kg

Volume of liquid displaced (v),

\frac{m}{v} = \frac{24}{1000} \\

\frac{m}{v} = 0.24m ^{3}

Now,

Since the volume of the wood is equal to the volume of water displaced, it is 0.024m^3

=====

Note:

=====

=> The volume of the wood below the water surface is the volume of water displaced.

=> Buoyant\: force = Weight\: of\: the \:displaced\: water.

8 0
3 years ago
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