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2 years ago

Ite

Physics

1 answer:

soldier1979 [14.2K]2 years ago

4 0

For the sound wave passing through regions of the ocean with varying density, longer wavelengths correspond to greater density of the water.

<h3>What is effect of density of a medium on wavelength of a wave?</h3>

The density of a medium is directly proportional to the wavelength of a wave.

The higher the density of the medium, the longer the wavelength of a wave.

Therefore, for a sound wave passing through regions of the ocean with varying density, longer wavelengths correspond to greater density of the water.

Learn more about density and wavelength at: brainly.com/question/9486264

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If the length of a building is 2 1 2 times the width and each dimension is increased by 7 ft, then the perimeter is 266 ft. Find

N76 [4]

Answer:

The original dimensions of the building is 95 ft × 38 ft.

Explanation:

Let the original length be 'l' and original width be 'w'.

Given:

Original length (l) = $2\frac{1}{2}\times original\ width$

Original width = 'w'.

So, $l=2\frac{1}{2}w=\frac{5}{2}w$

Now, as per question:

Length and width is increased by 7 ft.

So, new length (l') = $l+7=\frac{5w}{2}+7$

New width (w') = $w+7$

New perimeter (P) = 266 ft

Perimeter is given as:

$P=2(l' +w')\\\\266=2(\frac{5w}{2}+w)\\\\\frac{266}{2}=\frac{5w+2w}{2}\\\\266=7w\\\\w=\frac{266}{7}=38\ ft$

Therefore, original width = 38 ft.

Original length is, $l=\frac{5\times 38}{2}=\frac{190}{2}=95\ ft$

Hence, the original dimensions of the building is 95 ft × 38 ft.

7 0

4 years ago

A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

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