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fenix001 [56]
3 years ago
6

A force of 15 N will stretch a rubber band 10 cm ​(0.10 ​m). Assuming that​ Hooke's law​ applies, how far will an 18​-N force st

retch the rubber​ band? How much work does it take to stretch the rubber band this​ far?
Mathematics
1 answer:
Solnce55 [7]3 years ago
4 0

Answer: a) extension = 0.12m, b) work done = 2.16 J

Step-by-step explanation: according to hooke's law, provided that elastic limit is kept constant force (f) is proportional to extension (e).

F = ke, k = F / e.

F1/e1 = F2/e2

F1 = 15 N, e1 = 0.10m, F2= 18 N, e2 =?

15/0.10 = 18/e2

e2 = 18 * 0.10/ 15

e2 = 1.8/15

e2 = 0.12m.

Work done at this extension = F2 * e2

Work done = 18 * 0.12 = 2.16 J

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<h3>Further explanation</h3>

There are several forms of transformation, including translation, dilation, rotation or reflection

There are 2 shapes of geometry:

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the same shape but different in size,

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Let point H(4,1) from polygon GFJIH , we rotate -90°

\left[\begin{array}{ccc}x'\\y'\end{array}\right]=\left[\begin{array}{ccc}0&1\\-1&0\end{array}\right]\left[\begin{array}{ccc}4-6\\1+1\end{array}\right]+\left[\begin{array}{ccc}6\\-1\end{array}\right]\\\\=\left[\begin{array}{ccc}0&1\\-1&0\end{array}\right]\left[\begin{array}{ccc}-2\\2\end{array}\right]+\left[\begin{array}{ccc}6\\-1\end{array}\right]\\\\=\left[\begin{array}{ccc}8\\1\end{array}\right]

2. translation (0,2)

\left[\begin{array}{ccc}8\\1\end{array}\right]+\left[\begin{array}{ccc}0\\2\end{array}\right]=\left[\begin{array}{ccc}8\\3\end{array}\right]

<h3 /><h3>Learn more</h3>

brainly.com/question/13715226

Keywords: transformation, translation, rotation

#LearnWithBrainly

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