Science, asked by temp8722388999, 7 months ago

A man 'X" is standing inside a lift which is starting from rest moving up with an acceleration g/2. After 2 seconds 'X' drops a particle, acceleration of particle as observed by 'X' will be (in m/s') g = 10 m's?

Answers

Answered by arsalan9330
1

Answer:

Lift is accelerating downwards, so acceleration of lift w.r.t. ground  a  

lg

​  

=a

Acceleration of ball w.r.t. ground  a  

bg

​  

=g

Acceleration of ball w.r.t. lift   a  

bl

​  

=a  

bg

​  

−a  

lg

​  

=g−a

Explanation:

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Answered by SmritiSami
0

The acceleration of the particle is 15m/s² downwards.

Given:-

Acceleration of lift = g/2

Time = 2sec

To Find:-

The acceleration of the particle.

Solution:-

We can simply calculate the acceleration of the particle by using the following process.

As

Acceleration of lift (a) = g/2

g = 10m/s²

Time (t) = 2sec

The acceleration of the particle (g') =?

Since lift and particle both are accelerating so we can find the acceleration of particle by using the relative acceleration approach.

acceleration of particle = g - acceleration of the lift

g' = - g - \frac{g}{2}

the liftg' =  - g -  \frac{g}{2} g' =  \frac{ - 2g - g}{2}

the liftg' =  - g -  \frac{g}{2} g' =  \frac{ - 2g - g}{2} g' =  \frac{ - 3g}{2}

the liftg' =  - g -  \frac{g}{2} g' =  \frac{ - 2g - g}{2} g' =  \frac{ - 3g}{2} g' =  \frac{ - 3 \times 10}{2}

the liftg' =  - g -  \frac{g}{2} g' =  \frac{ - 2g - g}{2} g' =  \frac{ - 3g}{2} g' =  \frac{ - 3 \times 10}{2} g' =  \frac{ - 30}{2}

the liftg' =  - g -  \frac{g}{2} g' =  \frac{ - 2g - g}{2} g' =  \frac{ - 3g}{2} g' =  \frac{ - 3 \times 10}{2} g' =  \frac{ - 30}{2} g' =  - 15

the liftg' =  - g -  \frac{g}{2} g' =  \frac{ - 2g - g}{2} g' =  \frac{ - 3g}{2} g' =  \frac{ - 3 \times 10}{2} g' =  \frac{ - 30}{2} g' =  - 15Hence, The acceleration of the particle is 15m/ downwards.

#SPJ2

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