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When you drop a 0.43 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 :

skyluke89

The magnitude of the Earth's acceleration is [tex]7.0\cdot 10^{-25} m/s^2[/tex]

Explanation:

First of all, we start by calculating the magnitude of the force exerted by the apple on the Earth. According to Newton's third law, this is equal to the force exerted by the Earth on the apple, which is the weight of the apple, given by:

[tex]F=mg[/tex]

where

m = 0.43 kg is the mass of the apple

[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity

Substituting,

[tex]F=(0.43)(9.8)=4.2 N[/tex]

Now we can find the Earth's acceleration by applying Newton's second law:

[tex]F=Ma[/tex]

where:

F = 4.2 N is the net force exerted by the apple on the Earth

[tex]M=5.98\cdot 10^{24} kg[/tex] is the mass of the Earth

a is the Earth's acceleration

And solving for a, we find:

[tex]a=\frac{F}{M}=\frac{4.2}{5.98\cdot 10^{24}}=7.0\cdot 10^{-25} m/s^2[/tex]

Learn more about Newton laws of motion:

brainly.com/question/3820012

brainly.com/question/11411375

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