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EleoNora [17]
3 years ago
7

How are intensity sound and energy related

Physics
1 answer:
zubka84 [21]3 years ago
8 0
Sound is a form of energy in that it consists fluctuations of air pressure . The speed of the fluctuations is measured in cycles per second or Hertz (HZ)

Intensity is how large the fluctuations are, also known as amplitude and for the sound the unit is decibels of sonic pressure level (dB SPL)
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PLS HELP
Ahat [919]
<h3>B. True</h3>

"This was the idea that non-living objects can give rise to living organisms."

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Which substance is a greenhouse gas that is emitted by human activities, is the most widespread, and remains in the air the long
e-lub [12.9K]

Answer:

I am not really sure, but it is probably Carbon Dioxide

Explanation:

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3 years ago
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Question 5 of 10
jonny [76]

Answer:

A. The particle model, because only high-energy frequencies of light  can remove electrons .

Explanation:

Each photon of blue light has higher energy than each photon of red light has  . So when each photon strikes each electron , it gets ejected . But the photon of red light has not sufficient energy to eject electron . Once the photon of red light strikes the electron , the energy is wasted off . Energy of photon can not be accumulated . Thus photon behaves like particle .

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3 years ago
The current in a coil with a self-inductance of 1 mH is 2.8 A at t = 0, when the coil is shorted through a resistor. The total r
pogonyaev

Given:

L = 1 mH = 1\times 10^{-3} H

total Resistance, R = 11 \Omega

current at t = 0 s,

I_{o} = 2.8 A

Formula used:

I = I_{o}\times e^-{\frac{R}{L}t}

Solution:

Using the given formula:

current after t = 0.5 ms = 0.5\times 10^{-3} s

for the inductive circuit:

I = 2.8\times e^-{\frac{11}{1\times 10^{-3}}\times 0.5\times 10^{-3}}

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5 0
2 years ago
A ball with an initial velocity of 8.00 m/s rolls up a hill without slipping. (a) Treating the ball as a spherical shell, calcul
GrogVix [38]

Answer:

Part i)

h = 5.44 m

Part ii)

h = 3.16 m

Explanation:

Part i)

Since the ball is rolling so its total kinetic energy in this case will convert into gravitational potential energy

So we have

\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh

here we know that for spherical shell and pure rolling conditions

v = R \omega

I = \frac{2}{3}mR^2

\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2)(\frac{v^2}{R^2}) = mgh

\frac{5}{6}mv^2 = mgh

h = \frac{5v^2}{6g}

h = \frac{5(8^2)}{6(9.81)} = 5.44 m

Part b)

If ball is not rolling and just sliding over the hill then in that case

\frac{1}{2}mv^2 = mgh

h = \frac{v^2}{2g}

h = \frac{8^2}{2(9.81)} = 3.16 m

3 0
3 years ago
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