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hjlf
3 years ago
10

Imagine that Kevin can instantly transport himself between Planet X and Planet Y. Which statement could be said about Kevin in t

his situation? Kevin’s weight may change between Planet X and Planet Y. Kevin’s volume may change between Planet X and Planet Y. Kevin’s mass may change between Planet X and Planet Y. Kevin’s matter may change between Planet X and Planet Y.
Chemistry
1 answer:
Temka [501]3 years ago
4 0

Answer:

Kevin’s weight may change between Planet X and Planet Y.

Explanation:

Mass is the amount of matter contained in an object. Gravity is Earth’s pull of a body towards its centre.  

Weight (W) is product of the mass (m) of the object and the force of gravity (g) acting on the object.  

W = mg  

The force of gravity may change between Plant X and Planet Y, and thus Kevin’s weight may change between Planet X and Planet Y.  

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PLS HELP What is the mass of hydrogen atoms that is measured at 54 u?
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The mass of hydrogen atoms that is measured at 54 u given the relationship is 89.64×10¯²⁴ g

<h3>Conversion scale </h3>

1 u = 1.66×10¯²⁴ g

<h3>How to determine the mass of hydrogen atoms </h3>
  • Mass of Hydrogen (u) = 54 u
  • Mass of Hydrogen (g) =?

1 u = 1.66×10¯²⁴ g

Therefore

54 u = 54 × 1.66×10¯²⁴ g

54 u = 89.64×10¯²⁴ g

Thus, the mass of the hydrogen atoms measured at 54 u is 89.64×10¯²⁴ g

Learn more about conversion:

brainly.com/question/2139943

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"A sphere of radius 0.50 m, temperature 27oC, and emissivity 0.85 is located in an environment of temperature 77oC. What is the
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Explanation:

It is known that formula for area of a sphere is as follows.

                     A = 4 \pi r^{2}

                        = 4 \times 3.14 \times (0.50 m)^{2}

                        = 3.14 m^{2}

    T_{a} = (27 + 273.15) K = 300.15 K

          T = (77 + 273.15) K = 350.15 K

Formula to calculate the net charge is as follows.

             Q = esA(T^{4} - T^{4}_{a})

where,    e = emissivity = 0.85

               s = stefan-boltzmann constant = 5.6703 \times 10^{-8} Wm^{-2} K^{-4}

                A = surface area

Hence, putting the given values into the above formula as follows.

                 Q = esA(T^{4} - T^{4}_{a})

                     = 0.85 \times 5.6703 \times 10^{-8} Wm^{-2} K^{-4} \times 3.14 \times ((350.15)^{4} - (300.15)^{4})

                     = 1046.63 W

Therefore, we can conclude that the net flow of energy transferred to the environment in 1 second is 1046.63 W.

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