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Pavel [41]
3 years ago
11

Yosef is playing with different kinds of rubber bands. Some are very narrow, and some are quite wide. Yosef is curious about the

rubber bands and develops this scientific question: Does the width of a rubber band affect how easily it can be stretched? He decides to develop a hypothesis to test this scientific question. What could Yosef’s hypothesis be?
Physics
2 answers:
Cerrena [4.2K]3 years ago
5 0

Answer:

Yosef hypothesis could be stretching of rubber band depends on rubber band's width. It is difficult to stretch a wider rubber band in comparison to a narrow band.  

Explanation:

Width of rubber band affect how easily it can be stretched. It is found that it is difficult to to stretch a wider rubber band in comparison to a narrow band because when rubber band is narrow less molecules will be there along its width and hence less restoring force will be there so rubber can be easily stretched. On the other hand when the rubber band is wider it means more molecules  are there along its width and hence more restoring force will be there while stretching so it will be difficult to stretch a wider rubber band.

Romashka-Z-Leto [24]3 years ago
3 0

Answer:

If the width of a rubber band is increased, then it will be more difficult to stretch because more force will be needed to stretch it.

Explanation:

hope this helps

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Answer:

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Explanation:

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We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or \frac{\pi}{4} radian.

From trigonometry, \sin A =\cos B if A and B are complementary.

At t = 0, x = 3.5

3.5 = A\cos(\omega \times0)

A =3.5

So

x = 3.5\cos(\omega t)

At t = 0.12, x = 1.5

1.5 = 3.5\cos(0.12\omega)

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The period, T, is related to \omega by

T = \dfrac{2\pi}{\omega} = \dfrac{2\times3.14}{9.4}=0.67

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