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Sagot :
Answer:
F - M a force exerted by scales on student
M a = M (9.8 + 4.9) m/s^2 upwards chosen as positive
a = 1.5 g net acceleration of student due to force of scales
W =M g weight of student (actual weight)
Wapp = M 1.5 * g apparent weight (on scales) of student
In an elevator which is accelerating upwards with 4.9 m/[tex]s^{-2}[/tex], Mike's weight will be 1.5 times the actual weight.
In the question, it is given that the lift is accelerating upwards with 4.9 m/[tex]s^{-2}[/tex]
- Let Mike's weight be "m", "g" is acceleration due to gravity and "a" be the upward acceleration
- The upward acceleration of lift = 4.9 m/[tex]s^{-2}[/tex]
- As the lift is accelerating upwards, the apparent weight (R) = m (g+a)
- Apparent weight R = m*(9.8+4.9) N
- Apparent weight R = m*(9.8+4.9)/9.8 Kgf
- Apparent weight R = m*1.5
Hence, the apparent weight will be 1.5 times the real weight.
To know more about the relationship between an object accelerating upwards and the weight of the object, refer:
https://brainly.com/question/2293244
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