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Anna007 [38]
3 years ago
9

The contraption below has the following

Physics
1 answer:
AnnyKZ [126]3 years ago
4 0

Answer:

117.72 N

Explanation:

The given parameters are;

The mass m₁ = 2.0 × 10³ kg

The mass m₂ = 4.4 × 10² kg

The mass of the man, m₃ = 6.0 × 10 kg

The condition of the interaction of the surfaces = Frictionless surfaces

The

The tension in the string = The downward force = The weight of (m₂ + m₃) = (m₂ + m₃) × g

Let <em>a</em> represent the acceleration of the connected masses due to the weight of m₂, and m₃, we have;

(m₁ + m₂ + m₃) × a = (m₂ + m₃) × g

∴ a = (m₂ + m₃) × g/(m₁ + m₂ + m₃)

Which gives;

a = (4.4 × 10²+ 6.0 × 10) × 9.81/(2.0 × 10³+ 4.4 × 10²+ 6.0 × 10) = 1.962

The downward acceleration, a = 1.962 m/s²

The apparent weight of the man = The mass of the man, m₃ × The acceleration, <em>a</em>

∴ The apparent weight of the man = 6.0×10 kg ×1.962 m/s² = 117.72 N

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Answer:

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Solution:

As per the question:

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n = \frac{d}{d'} = \frac{314.99}{2\pi \times 25} = 2

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