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Vladimir79 [104]
3 years ago
5

A metal rod with a length of 25.0 cm lies in the xy-plane and makes an angle of 37.5 ∘ with the positive x-axis and an angle of

52.5 ∘ with the positive y-axis. The rod is moving in the +x-direction with a speed of 6.80 m/s . The rod is in a uniform magnetic field B⃗ =(0.130T)i^−(0.290T)j^−(0.0800T )k^.
What is the magnitude of the emf induced in the rod?
Physics
1 answer:
pashok25 [27]3 years ago
5 0
We are given with the following:
L = 25 cm
θx = <span>37.5</span>°<span>
</span>θy = 52.5°
vx = 6.8 m/s
B = 0.13i - 0.29j - 0.08k

And we are asked for the emf induced in the rod
E = B L v
Substitute the values for B and L into the equation. For v, use the given velocity along x and the angles to convert it to vector form. Then, solve for E.
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Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the
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Answer:

 K_a = 8,111 J

Explanation:

This is a collision exercise, let's define the system as formed by the two particles A and B, in this way the forces during the collision are internal and the moment is conserved

initial instant. Just before dropping the particles

          p₀ = 0

final moment

          p_f = m_a v_a + m_b v_b

          p₀ = p_f

          0 = m_a v_a + m_b v_b

tells us that

          m_a = 8 m_b

         

           0 = 8 m_b v_a + m_b v_b

           v_b = - 8 v_a                    (1)

indicate that the transfer is complete, therefore the kinematic energy is conserved

starting point

           Em₀ = K₀ = 73 J

final point. After separating the body

          Em_f = K_f = ½ m_a v_a² + ½ m_b v_b²

           K₀ = K_f

           73 = ½ m_a (v_a² + v_b² / 8)

           

we substitute equation 1

           73 = ½ m_a (v_a² + 8² v_a² / 8)

           73 = ½ m_a (9 v_a²)

           73/9 = ½ m_a (v_a²) = K_a

            K_a = 8,111 J

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3 years ago
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