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Nezavi [6.7K]
3 years ago
14

Two 3 m long conductors are separated from each other by 5 mm and carry a current of 10 A dc. Calculate the force that one condu

ctor exerts on the other.
Physics
1 answer:
makkiz [27]3 years ago
7 0

Answer:

0.012 N

Explanation:

The formula to apply is that adopted from the Ampere law which is;

F= μ* I₁*I₂*l /2πd   where

F is force that one conductor exerts on the other.

μ = magnetic permeability of free space = 4π×10⁻⁷ T. m/A

I₁ = current in conductor one=10 A

I₂ = current in conductor two= 10 A

l= length of conductor= 3 m

d= distance between the conductors = 5 mm = 0.005 m

Applying the values in the equation

F= 4π×10⁻⁷ *10*10*3 / 2π*0.005

F= 6 * 10⁻⁵ / 0.005

F=0.012 N

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Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic colli
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Answer:

a)  v = 3,843 m / s, b)  46.7º  North- East

Explanation:

Moment is a vector quantity, so one of the best ways to solve this problem is to solve each component separately.

The system is formed by the two vehicles so that the moment is preserved during the crash

Direction to the East    

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          p₀ = mₐ vₐ₀

final insttne. After the crash

          p_f = (mₐ + m_b) vₓ

         p₀ = p_f

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         vₓ = \frac{m_a}{m_a + m_b} \ v_{ao}

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          vₓ = \frac{16.7}{16.7 + 29.3} \ 7.26

          vₓ = 2,636 m / s

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initial   p₀ = m_b v_{bo}

final     p_f = (mₐ + m_b) v_y

          p₀ = p_f

          m_b v_{bo} = (mₐ + m_b) v_y

          v_y = \frac{m_b}{m_a+m_b} \ v_{bo}

let's calculate

          v_y = \frac{29.3}{16.7 + 29.3} \ 4.39

          v_y = 2.796 m / s

the final speed of the two two vehicles is

          v = (2,636 i ^ + 2,796 j ^) m / s

a) the magnitude of the velocity

let's use the Pythagorean theorem

       v = \sqrt{v_x^2 + v_y^2}

      v = \sqrt{2.636^2 + 2.796^2}

      v = 3,843 m / s

b) let's use trigonometry to find the direction

      tan θ = v_y / vₓ

      θ = tan⁻¹ v_y / vₓ

      θ = tan⁻¹ (2,796 / 2,636)

      θ = 46.7º

This direction is 46.7º  North East

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