Nitha Mulungu
May 25
What is the apparent weight of a 50kg person in a lift that is acceleration upward at 1.5m/s^2?
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ChatGPT
May 25
The apparent weight of a person in an elevator is equal to the sum of their actual weight and the force exerted on them by the acceleration of the elevator.
The actual weight of the person is 50kg * 9.8m/s^2 = 490N.
The force exerted on the person by the acceleration of the elevator is equal to the mass of the person multiplied by the acceleration of the elevator, which is 50kg * 1.5m/s^2 = 75N.
Therefore, the apparent weight of the person in the elevator is 490N + 75N = 565N.
The actual weight of the person is 50kg * 9.8m/s^2 = 490N.
The force exerted on the person by the acceleration of the elevator is equal to the mass of the person multiplied by the acceleration of the elevator, which is 50kg * 1.5m/s^2 = 75N.
Therefore, the apparent weight of the person in the elevator is 490N + 75N = 565N.
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Google Gemini AI
May 25
Apparent weight = Actual weight + Buoyant force - Inertial force
Buoyant force = Density of air - Volume of displaced air - g
Inertial force = Mass - Acceleration
∴ Apparent weight = 50 - 9.8 + 0 - (50 - 1.5)
= 490 - 75
= 415 N
Buoyant force = Density of air - Volume of displaced air - g
Inertial force = Mass - Acceleration
∴ Apparent weight = 50 - 9.8 + 0 - (50 - 1.5)
= 490 - 75
= 415 N
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