A70 kg man's arm, including the hand, can be modeled as bio 75-cm-long uniform rod with a mass of 3.5 kg. when the man raises both his arms, from hanging down to straight up, by how much does he raise his center of mass?

Given,

The weight of the man is 70 kg.

The mass of the rod is 3.5 kg.

Computation:

Raising the arm, therefore lifts the center of mass of the arm through, firstly 37.5cm to get the arm straight out in front of you, then another 37.5cm to get the arm straight up.

Total is 7cm.

Finally, work out the ratio of the mass of the arms to the total mass of the person

So, the raised center of mass is,

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  Δy = 7.1 cm

Explanation:

The center of mass of a body is defined

            = 1 /M ∑i

Where M is the total mass of the body, m mass of each part and ‘y’ height

Let's apply this equation to our case

We locate the reference system on the shoulders

The height of the arms is at its midpoint

            y = -75/2 = 37.5 cm

With arms down

            = 1/70 (63 y₀ - 3.5 37.5 - 3.5 37.5)

            = 1/70 (63 y)₀ - 7 37.5)

With arms up

          ’= 1/70 (63 y₀ + 3.5 y + 3.5 y)

          ’= 1/70 (63y₀ + 7 35.5)

let's subtract the two equations

        ’ - = 1/70 2 (7 35.5)

         Δy = ’ - = 2 7 35.5 / 70

         ΔY = 7.1 cm

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

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