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OverLord2011 [107]
3 years ago
14

A ship can carry more load than aeroplane of same size. why?​

Physics
1 answer:
stepladder [879]3 years ago
8 0

Answer:

density of water is much greater than density of air

Explanation:

when body is in fluid it carries weight which is always equal to upthrust

upthrust is given as

upthrust= density of fluid ×strength of gravitational field g × volume of body in fluid

as volume of ship and aeroplane is same and strength of gravitational field or acceleration due to gravity is same hence upthrust depens upon density of fluid .

density of air is about 1.2 kg/m^3  and density of water is 1000kg/m^3 hence ship carries more load

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Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and th
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Complete Question

Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and thus our bodies) as three isotopes: Potassium-39, Potassium-40, and Potassium-41. Their current abundances are 93.26%, 0.012% and 6.728%. A typical human body contains about 3.0 grams of Potassium per kilogram of body mass. 1. How much Potassium-40 is present in a person with a mass of 80 kg? 2. If, on average, the decay of Potassium-40 results in 1.10 MeV of energy absorbed, determine the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body. Assume an RBE of 1.2. The half-life of Potassium-40 is 1.28 * 10^9years.

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The potassium-40 present in 80 kg is  Z = 0.0288 *10^{-3}\ kg

The effective dose absorbed per year is  x = 2.06 *10^{-24} per year

Explanation:

From the question we are told that

      The mass of potassium in 1 kg of human body is m =  3g= \frac{3}{1000} =  3*10^{-3} \ kg

      The mass of the person is M = 80 \ kg

       The abundance of Potassium-39 is   93.26%

        The abundance of Potassium-40 is   0.012%

         The abundance of Potassium-41 is   6.78 %

         The energy absorbed is  E =  1.10MeV = 1.10 *10^{6} * 1.602 *10^{-19} = 1.7622*10^{-13} J

Now  1 kg of human body contains       3.0*10^{-3}\ kg of  Potassium

So      80 kg of human body contains      k kg of  Potassium

=>   k = \frac{ 80 * 3*10^{-3}}{1}

     k = 0.240\  kg

Now from the question potassium-40 is  0.012% of the total  potassium so

     Amount of potassium-40  present is mathematically represented as

            Z = \frac{0.012}{100}  * 0.240

            Z = 0.0288 *10^{-3}\ kg

The effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is mathematically evaluated as

           D =  \frac{E}{M}

Substituting values

          D =  \frac{1.7622*10^{-13}}{80}

            D =  2.2*10^{-15} J/kg

Converting to Sieverts

We have

           D_s = REB * D

           D_s = 1.2 * 2.2 *10^{-15}

           D_s =  2.64 *10^{-15}

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     for half-life (1.28 *10^9 \ years)  the dose is  2.64 *10^{-15}

     Then for 1  year the dose would be  x

=>         x = \frac{2.64 *10^{-15}}{1.28 * 10^9}

             x = 2.06 *10^{-24} per year      

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