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yaroslaw [1]
4 years ago
6

A 4.2 m board with a mass of 19 kg is pivoted at its center of gravity. A helium balloon attached 0.24 m from the left end of th

e board produces an upward force of 7.1 N. A 3.5 kg book is placed 0.74 m from the left end of the board, and another book of 1.7 kg is placed 0.77 m from the right end of the board. Find the torque on the board and the direction of rotation. (Hint: add the torques.)
Physics
1 answer:
storchak [24]4 years ago
6 0

Answer:

The torque on the board = 11.8 Nm in the anti-clockwise direction

Explanation:

Please find attached the explanation to the answer given.

Download docx
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sammy [17]

Answer:

1.24 C

Explanation:

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The are , A of the circular loop is πD²/4 where D = diameter of circular loop = 16.7 cm = 16.7 × 10⁻²m

So, ε = -ΔΦ/Δt = -AΔB/Δt= -πD²/4 × -0.750 T/Δt = 0.750πD²/4Δt.

Also, the induced emf ε = iR where i = current in the coil and R = resistance of wire = ρl/A where ρ = resistivity of copper wire =1.68 × 10⁻⁸ Ωm, l = length of wire = πD and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m.

So, ε = iR = iρl/A = iρπD/πd²/4 = 4iρD/d²

So,  4iρD/d² = 0.750πD²/4Δt.

iΔt = 0.750πD²/4 ÷ 4iρD/d²

iΔt = 0.750πD²d²/16ρ.

So the charge Q = iΔt

= 0.750π(Dd)²/16ρ

= 0.750π(16.7 × 10⁻²m 2.25 × 10⁻³ m)²/16(1.68 × 10⁻⁸ Ωm)

= 123.76 × 10⁻² C

= 1.2376 C

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Answer:

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