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antiseptic1488 [7]
3 years ago
7

A 70 kg box undergoes a horizontal acceleration of 3.0 m/s^2 on a level surface when pulled by a 130 N force, What is the coeffi

cient of kinetic friction between the box and the surface?
Physics
1 answer:
elixir [45]3 years ago
7 0

I think there's a typo in the question...

Vertically, the box is in equilibrium, so its weight and the normal force (magnitudes <em>w</em> and <em>n</em>, respectively) are such that

<em>n</em> + (-<em>w</em>) = 0

The box has mass 70 kg, and assuming gravitational acceleration with magnitude <em>g</em> = 9.80 m/s², it has a weight of

<em>w</em> = (70 kg) <em>g</em> = 686 N

and hence

<em>n</em> = 686 N

Horizontally, the box is accelerated 3.0 m/s², so the net force acting on it is

∑ <em>F</em> = (70 kg) (3.0 m/s²) = 210 N

and the only forces acting in this dimension are the pulling force with magnitude 130 N and the friction force with magnitude <em>f</em> so that

130 N - <em>f</em> = 210 N

The friction force is proportional to the normal force by a factor of <em>µ</em>, the coefficient of kinetic friction, so that

<em>f</em> = <em>µ</em> <em>n</em>

and so we have

130 N - <em>µ</em> (686 N) = 210 N   →   <em>µ</em> ≈ -0.117

but <em>µ</em> can't be negative!

The problem is that the pulling force should have a magnitude larger than that of the net force, so either the given mass, acceleration, or pulling force are incorrect.

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If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle
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This question is incomplete, the complete question is;

The electric force due to a uniform external electric field causes a torque of magnitude 20.0 × 10⁻⁹ N⋅m on an electric dipole oriented at 30° from the direction of the external field. The dipole moment of the dipole is 7.5 × 10⁻¹² C⋅m.

What is the magnitude of the external electric field?

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle?

Answer:

- the magnitude of the external electric field is 5333.3 N/C

- the magnitude of the charge on each particle is 3.0 × 10⁻¹² C  ≈ 3 nC

Explanation:

Given that;

Torque = 20.0 × 10⁻⁹ N⋅m

dipole moment = 7.5 × 10⁻¹²

∅ = 30°

The moment T of restoring couple is;

T = PEsin∅

E = T/Psin∅

we substitute

E = 20.0 × 10⁻⁹ N⋅m / (7.5 × 10⁻¹²) sin(30°)

E = 20.0 × 10⁻⁹ / 3.75 × 10⁻¹²

E =  5333.3 N/C

Therefore, the magnitude of the external electric field is 5333.3 N/C

The dipole moment is given by the expression;

p = ql

q = p / l

given that l = 2.5 mm = 0.0025 m

we substitute

q = 7.5 × 10⁻¹² / 0.0025

q = 3.0 × 10⁻¹² C ≈ 3 nC

Therefore, the magnitude of the charge on each particle is 3.0 × 10⁻¹² C ≈ 3 nC

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