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Ulleksa [173]
3 years ago
9

Two 4.0 kg masses are 1.0 m apart on a frictionless table. Each has 1.0 μC of charge.What is the magnitude of the electric force

on one of the masses?
Physics
1 answer:
xeze [42]3 years ago
7 0

Coulomb's law:

  Force = (<span>8.99×10⁹ N m² / C²<span>) · (charge₁) · (charge₂) / distance²

            = (</span></span><span>8.99×10⁹ N m² / C²<span>) (1 x 10⁻⁶ C) (1 x 10⁻⁶ C)  / (1.0 m)²

            = (8.99×10⁹ x 1×10⁻¹² / 1.0)  N

            =      8.99×10⁻³  N

            =        0.00899 N repelling.

Notice that there's a lot of information in the question that you don't need.
It's only there to distract you, confuse you, and see whether you know
what to ignore.

-- '4.0 kg masses';  don't need it. 
   Mass has no effect on the electric force between them.

-- 'frictionless table';  don't need it.
   Friction has no effect on the force between them,
only on how they move in response to the force.
</span></span>
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3 years ago
One uniform ladder of mass 30 kg and 10 m long rests against a frictionless vertical wall and makes an angle of 60o with the flo
yuradex [85]

Answer:

   μ = 0.37

Explanation:

For this exercise we must use the translational and rotational equilibrium equations.

We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive

let's write the rotational equilibrium

           W₁  x/2 + W₂ x₂ - fr y = 0

where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances

             cos 60 = x / L

where L is the length of the ladder

              x = L cos 60

            sin 60 = y / L

           y = L sin60

the horizontal distance of man is

            cos 60 = x2 / 7.0

            x2 = 7 cos 60

we substitute

         m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0

         fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60

let's calculate

         fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)

         fr = (735 + 2450) / 8.66

         fr = 367.78 N

the friction force has the expression

         fr = μ N

write the translational equilibrium equation

         N - W₁ -W₂ = 0

         N = m₁ g + W₂

         N = 30 9.8 + 700

         N = 994 N

we clear the friction force from the eucacion

        μ = fr / N

        μ = 367.78 / 994

        μ = 0.37

3 0
3 years ago
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