contestada

When you drop a 0.44 kg apple, Earth exerts a force on it that accelerates it at 9.8 m/s 2 toward the earth’s surface. According to Newton’s third law, the apple must exert an equal but opposite force on Earth. If the mass of the earth 5.98 × 1024 kg, what is the magnitude of the earth’s acceleration toward the apple? Answer in units of m/s 2 .

Respuesta :

Hania12

Answer:

                [tex]7.217*10^{-25} }[/tex] m/s²

Explanation:

We can calculate by the following formulas.

       m =0.44 Kg

       g =9.81 m/s²

       M= 5.98*[tex]10^{24}[/tex] Kg

        F=m*g

         =0.44*9.81

         =4.316 N

      F=M*a

      a=[tex]\frac{F}{M}[/tex]

     a=[tex]\frac{4.316}{5.98^{24} }[/tex]

     a=[tex]7.217*10^{-25} }[/tex] m/s²

Otras preguntas