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3 years ago

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A 2.5-kg rock is dropped off a 32-m cliff and hits a spring, compressing it 57 cm. What is the spring constant? Round your answe

r to two significant figures.

2 answers:

Valentin [98]3 years ago

6 0

The answer is 4800 on the assignment. good luck

Alla [95]3 years ago

6 0

Answer:

The spring constant of the spring is $k=4.9\times 10^3\ N/m$ .

Explanation:

Mass of the rock is 2.5 kg

It is dropped off a 32 m cliff and hits a spring, compressing it 57 cm. It is required to find the spring constant.

When it falls form a height, it possess potential energy and when it hits a spring, the potential energy gets converted to elastic potential energy such that :

$mgh=\dfrac{1}{2}kx^2$

Here, k is spring constant

$k=\dfrac{2mgh}{x^2}\\\\k=\frac{2\cdot2.5\cdot10\cdot32}{0.57^{2}}\\\\k=4924.59\ N/m$

$k=4.9\times 10^3\ N/m$

So, the spring constant of the spring is $k=4.9\times 10^3\ N/m$ .

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Answer:

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To get the speed at witch the water raising at a given point we need to know the area it needs to fill at that point in the trough (the longitudinal section), which is given by the height at that point.

So we need to get the lenght of the sides for a height of 1 foot. Given the geometry of the trough, one side is the depth <em>d</em> and the other (lets call it <em>l</em>) is given by:

$l=\frac{3-2}{2}\,ft+2\,ft\\l=2.5\,ft$

since the difference between the upper and lower base is the increase in the base and we are only at halft the height.

Now we can calculate the longitudinal section <em>A</em> at that point:

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And the raising speed <em>v </em>of the water is given by:

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7 0

3 years ago

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Answer:

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sladkih [1.3K]

Explanation:

Given that,

Length of gold wire, l = 4 m

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3 years ago

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jeka57 [31]

Answer:

The answer is below

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Answer:

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