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Sam, the stunt pilot, is pulling out of a dive by guiding the airplane into a vertical circle with radius 390 m. At the bottom of the dive, the speed of the airplane is 121 m/s. What is Sam's apparent weight of the pilot at that point if his mass is 76.0 kg?

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Final answer:

Sam's apparent weight at the bottom of the dive, where the radius of the circle is 390 m and his speed is 121 m/s, is approximately 3602.4 newtons.

Step-by-step explanation:

The question asks to calculate Sam's apparent weight at the bottom of a dive if he is piloting an airplane in a vertical circle with a radius of 390 m and with a speed of 121 m/s.

To find this, we use the concept of circular motion.

The pilot's apparent weight is the force he feels from the seat, which is equal to the normal force.

At the bottom of the circle, the apparent weight is the centripetal force needed to keep the pilot moving in a circle plus the force of gravity.

The formula used is apparent weight = mass · (centripetal acceleration + g), where centripetal acceleration is v²/r and g is the acceleration due to gravity (9.8 m/s²).

Plugging in Sam's mass (76.0 kg), speed (121 m/s), and radius (390 m) into the formula, we calculate the centripetal acceleration as (121 m/s)² / 390 m, which equals approximately 37.6 m/s².

The apparent weight will be 76.0 kg · (37.6 m/s² + 9.8 m/s²) = 76.0 kg · 47.4 m/s² = 3602.4 N.

So Sam's apparent weight at the bottom of the dive is approximately 3602.4 newtons.

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