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krek1111 [17]
3 years ago
9

When a 4.60-kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.30 cm. (a) If

the 4.60-kg object is removed, how far will the spring stretch if a 1.50-kg block is hung on it? cm (b) How much work must an external agent do to stretch the same spring 4.00 cm from its unstretched position?

Physics
1 answer:
Alex787 [66]3 years ago
6 0

Answer:

a) y = 0.0075 m

b) W = 1.569 J

Explanation:

See attachment for the solution

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Answer:

The number of atoms is N = 4.37*10^{22} \ atoms

Explanation:

From the question we are told that

                 The mass of coin  is m_n = 4.2g

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                        N = \frac{Mass \ of Nickel\ coin}{Molar \ mass\ of nickel }  * No\ of\atoms \ in \ \ one\  mole\ of\ nickel

                           = \frac{4.2}{57.8}* 6.02*10^{23}

                           =4.37 *10^{22} atoms

                 

     

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