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3 years ago

In Science, what is a model?

Physics

2 answers:

abruzzese [7]3 years ago

8 0

I'm not sure about the "in nature" part but i think its A

Citrus2011 [14]3 years ago

8 0

Pretty sure it’s A

Hope this helps

-AaronWiseIsBae

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which principles of training refers o placing increased demands on the body? A. Specificity B. cross-training c.type D.overload

Anuta_ua [19.1K]

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5 0

3 years ago

A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound

NemiM [27]

Answer:

f3 = 102 Hz

Explanation:

To find the frequency of the sound produced by the pipe you use the following formula:

$f_n=\frac{nv_s}{4L}$

n: number of the harmonic = 3

vs: speed of sound = 340 m/s

L: length of the pipe = 2.5 m

You replace the values of n, L and vs in order to calculate the frequency:

$f_{3}=\frac{(3)(340m/s)}{4(2.5m)}=102\ Hz$

hence, the frequency of the third overtone is 102 Hz

8 0

3 years ago

Which is a more objective measurement, sound intensity or loudness, and why? 1. sound intensity is exactly same as loudness. 2.

fgiga [73]

The correct answer is:
<span>2. sound intensity is a more objective and physical attribute of a sound wave because loudness can vary from person to person

indeed, sound intensity is a measurable quantity, and so it is objective, while loudness is the subjective perception of the sound level, so it varies from person to person.</span>

7 0

3 years ago

Read 2 more answers

Does frequency affect wavelength?

Vinil7 [7]

Answer:

yeah

Explanation:

as wavelength increases frequency decreases and it goes the same for the opposite way

6 0

3 years ago

A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.4 m/s. When the pipe narrows to one half its

Pavel [41]

Answer:

$v_2 = 5.6 \ m/s$

$P_2 = 80600 \ Pa$

Explanation:

From the question we are told that

The pressure of the water in the pipe is $P_1= 110 \ kPa = 110 *10^{3 } \ Pa$

The speed of the water is $v_1 = 1.4 \ m/s$

The original area of the pipe is $A_1 = \pi \frac{d^2 }{4}$

The new area of the pipe is $A_2 = \pi * \frac{[\frac{d}{2} ]^2}{4} = \pi * \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}$

Generally the continuity equation is mathematically represented as

$A_1 * v_1 = A_2 * v_2$

Here $v_2$ is the new velocity

$\pi * \frac{d^2}{4} * 1.4 = \pi * \frac{d^2}{16} * v_2$

=> $\frac{d^2}{4} * 1.4 = \frac{d^2}{16} * v_2$

=> $d^2 * 1.4 = \frac{d^2}{4} * v_2$

=> $1.4 = 0.25 * v_2$

=> $v_2 = 5.6 \ m/s$

Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the Bernoulli's equation for constant height to calculate the pressure

This is mathematically represented as

$P_1 + \frac{1}{2} * \rho * v_1 ^2 = P_2 + \frac{1}{2} * \rho * v_2 ^2$

Here $\rho$ is the density of water with value $\rho = 1000 \ kg /m^3$

$P_2 = P_1 + \frac{1}{2} * \rho [ v_1^2 - v_2^2 ]$

=> $P_2 = 110 *10^{3} + \frac{1}{2} * 1000 * [ 1.4 ^2 - 5.6 ^2 ]$

=> $P_2 = 80600 \ Pa$

4 0

2 years ago