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3 years ago

The speed of sound in ice, water, and steam is shown. What best explains the speed of sound in different states of matter?

2 answers:

5 0

The speed of sound is greater in ice (4000 m/s), then in water (1500 m/s), then in air (340 m/s). The explanation for this is the differente state of the matter in the three cases.

In fact, sound waves travel faster in solids (like ice), then in liquids (like water), then in gases (like air). This is because the speed of the sound wave depends on the density of the medium: the greater the density, the faster the sound wave. This can be easily understood by thinking at how a sound wave propagates: a sound wave is a vibration of molecules, which is transmitted throughout the medium by collision of the molecules. Therefore, the smaller the spacing between the molecules (such as in solids), the more efficient is the propagation, and so the sound wave is faster. On the contrary, there is a large spacing between molecules in gases (such as in the air), so there are less collisions between the molecules and so the wave is not transmitted efficiently, and so it has less velocity.

nataly862011 [7]3 years ago

4 0

Answer:

A i think

Explanation:

on edge

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A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (

SIZIF [17.4K]

The solution to the questions are given as

$t=40.39 \mathrm{sec}$
$\varepsilon &=(0.12v)e^{0.057t}$
the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}$

$\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$$

$\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}$

$\varepsilon &=(0.12v)e^{0.057t}$

(b) $Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$$

$\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}$

In conclusion, the direction of the induced current will be Counterclockwise.

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