Respuesta :

Answer:

[tex]a=0.125\ m/s^2[/tex]

Explanation:

Given that,

Mass, m = 40 kg

The net force with which Joyce pushes Efua is 5 N.

We need to find Efua’s resulting acceleration. Let it is equal to a. Net force is given by :

F = ma

[tex]a=\dfrac{F}{m}\\\\a=\dfrac{5\ N}{40\ kg}\\\\a=0.125\ m/s^2[/tex]

So, the acceleration is [tex]0.125\ m/s^2[/tex].

snehashish65

The resulting acceleration of Efua is  [tex]0.125 \;\rm m/s^{2}[/tex].

Given data:

The mass of Efua is, m = 40.0 kg.

The magnitude of applied force by Joyce is, F = 5.00 N.

The given problem is based on the Newton's Second law of motion. As per the Newton's second law, the applied force is equal to the product of mass and the acceleration of object. Therefore,

F = ma

a = F/m

Solving as,

a = 5.00 / 40

[tex]a = 0.125 \;\rm m/s^{2}[/tex]

Thus, we can conclude that the resulting acceleration of Efua is  [tex]0.125 \;\rm m/s^{2}[/tex].

learn more about the Newton's second law here:

https://brainly.com/question/19226427

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