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student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 834 N. (a) As the elevator moves up, the scale reading increases to 928 N. Find the acceleration of the elevator. 1.11 Correct: Your answer is correct. m/s2 (b) As the elevator approaches the 74th floor, the scale reading drops to 782 N. What is the acceleration of the elevator?

Respuesta :

Answer:

(a) 1.1 [tex]m/sec^2[/tex]

(b)  0.611[tex]m/sec^2[/tex]

Explanation:

We have given that when the elevator is at rest the scales reads 834 N

So W = 834 N

Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]

We know that W = mg

So [tex]m=\frac{W}{g}=\frac{834}{9.8}=85.1020kg[/tex]

(a)Now as the elevator moves upward so effective acceleration = g+a

Scale reading W = 928 N

Mass = 85.1020 kg

So [tex]g+a=\frac{928}{85.1020}=10.9045m/sec^2[/tex]

So  a = 10.9045 -9.8 = 1.1 [tex]m/sec^2[/tex]

(b) As the scale reading decreases effective acceleration = g-a

Scale reading = 782 N

So [tex]g-a=\frac{782}{85.1020}=8.1819m/sec^2[/tex]

So a =9.8 -9.1819 = 0.611[tex]m/sec^2[/tex]

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