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ICSE Class 9 Answered

Please answer the following questions with figure.
1.An elevator starts from rest with a constant upward acceleration.It moves 2 m in the first 0.6s.A passenger in the elevator is holding a 3kg package by a vertical string.The tension in the string during acceleration is:
2.A boy spots a ball that goes up past an open window.The ball,which passes very close to the window is in view of the boy for 0.25s while going up and the top to bottom height of the window is 2m.What is the ball upward velocity at the bottom of the window? 
3.A closed compartment containing a gas is moving with some acceleration in horizontal direction.Neglect effect of gravity.Then the pressure in the compartment is:
 
Asked by goyal.rishabh77 | 24 Sep, 2015, 06:00: PM
answered-by-expert Expert Answer
1. The acceleration of the elevator can be found out by using 2nd equation of motion:
Here, initial velocity of elevator, u = 0 m/s and distance travelled in time t= 0.6 sec is s = 2m
begin mathsize 12px style straight s space equals space ut space plus space 1 half at squared 2 space space equals space space 1 half straight a open parentheses 0.6 space straight s close parentheses squared rightwards arrow straight a space equals space 11.11 space straight m divided by straight s squared end style
Initially when the lift is at rest, the weight is acting downwards = mg
Acceleration of lift is in the upward direction = a
Therefore, the relative acceleration when the lift moves upwards is
a1 = (g+a)
    = (9.8 + 11.11)
a1 = 20.91 m/s2
Tension in the string is
T = m × a1
T = 3 × 20.91
T = 62.73 N
 
Note: Please ask the other questions as a separate query.
Answered by Yashvanti Jain | 24 Sep, 2015, 07:01: PM

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