The contraption below has the following characteristics: m1 = 2.0 x 103 kg, m2 = 4.4 x 102 kg,
the man on top of m2 has a mass m = 6.0 x 101 kg,
and all surfaces are frictionless. What is the apparent
weight of the man?

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 a 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, a

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

= initial pressure

= final pressure = 2

= initial volume

= final volume = /4

= initial temperature = 20 c = 20 + 273 = 293 k

= final temperature = ?

using the equation

365.25 earth days are in the year of a light year