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mina [271]
1 year ago
7

A mass of 0.4 kg hangs motionless from a vertical spring whose length is 0.76 m and whose unstretched length is 0.41 m. Next the

mass is pulled down to where the spring has a length of 1.01 m and given an initial speed upwards of 1.6 m/s. What is the maximum length of the spring during the motion that follows?
Physics
1 answer:
Blizzard [7]1 year ago
6 0
<h3><u>Answer;</u></h3>

= 1.256 m

<h3><u>Explanation;</u></h3>

We can start by finding the spring constant  

F = k*y  

Therefore;  k = F/y = m*g/y

                               = 0.40kg*9.8m/s^2/(0.76 - 0.41)

                               = 11.2 N/m  

Energy is conserved  

Let A be the maximum displacement  

Therefore;  1/2*k*A^2 = 1/2*k*(1.20 - 0.41)^2 + 1/2*m*v^2  

Thus;  A = sqrt((1.20 - 0.55)^2 + m/k*v^2)

               = sqrt((1.20 -0.55)^2 + 0.40/9.8*1.6^2)

                = 0.846 m  

Thus; the length will be 0.41 + 0.846  = 1.256 m

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Answer:

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1 year ago
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Veseljchak [2.6K]

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Explanation:

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here we know that when hands are shake then it will have paint on it. so here due to hand shake the hands are in contact with charge particles and due to contact the electrons are transferred to the hand.

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SO here correct answer will be

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5 0
1 year ago
Read 2 more answers
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