English, asked by vaishu47, 1 year ago

The work done in stretching a spring of spring constant 100N/m through a length of 10 cm is

Answers

Answered by Misha20

Work done = 1/2kx^2
=1/2*100*(10*10^-2)^2
=0.5 J

Misha20: What are u saying

mudrabhandari1p4cb5t: this is. incorrect

vaishu47: yes

Misha20: Yes yes my mistake sorry

vaishu47: nothing happens

vaishu47: one more question

Misha20: yes

vaishu47: The work done in lifting a body of mass m to a height equal to the radius of earth is

Misha20: Work done = mgh/(1+h/R). Since the height is equal to the radius ie R therefore mgh/(1+R/R)= mgh/2

vaishu47: not understanding

Answered by mudrabhandari1p4cb5t

$\frac{1}{2} k {x}^{2} = 0.5 \times 100 \times ({10 \times {10}^{ - 2} )}^{2} = 0.5$

mudrabhandari1p4cb5t: you have to calculate it with the formula but question is about ratio so you can just solve it with the proportionality

vaishu47: ohh

vaishu47: ok

vaishu47: hiw to take a pic

mudrabhandari1p4cb5t: you have to write your question the click on ask the question and there option for camera will appear that you can post the pic of your question

vaishu47: not getting the option

vaishu47: y I am not getting

mudrabhandari1p4cb5t: how am I able to make you understand I can't send any pic in comments

vaishu47: ya

vaishu47: thank you

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