A piece of tape is pulled from a spool and lowered toward a 120-mg scrap of paper. Only when the tape comes within 8.0 mm is the electric force magnitude great enough to overcome the gravitational force exerted by Earth on the scrap and lift it.

Respuesta :

Answer:

[tex]1.176\times 10^{-3} N[/tex]

Upward

Explanation:

We are given that

Mass of scarp paper,[tex]m=120 mg=120\times 10^{-6}kg[/tex]

1mg=[tex]10^{-6} mkg[/tex]

Distance,d =8 mm=[tex]8\times 10^{-3} m[/tex]

Magnitude of electric force =[tex]F_E= w=mg[/tex]

Where [tex]g=9.8 m/s^2[/tex]

Substitute the values

[tex]F_E=120\times 10^{-6}\times 9.8[/tex]

[tex]F_E=1.176\times 10^{-3} N[/tex]

Gravitational force act in  downward direction.

The electric force acts in opposite direction and magnitude of electric force is equal to gravitational force.

Hence, the direction of electric force is upward.

The electric force magnitude will be "1.176 × 10⁻³ N" in upward direction. To understand the calculation, check below.

Electric and Gravitational force

According to the question,

Scrap paper's mass, m = 120 mg or,

                                      = 120 × 10⁻⁶ kg

Distance, d = 8.0 mm or,

                   = 8 × 10⁻³ m

Acceleration due to gravity, g = 9.8 m/s²

We know the formula,

Electric force, [tex]F_E[/tex] = w or,

                                = mg

By substituting the values,

                                = 120 × 10⁻⁶ × 9.8

                                = 1.176 × 10⁻³ N (upward)

Thus the above approach is correct.

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