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abruzzese [7]
3 years ago
12

The intensity of a sound wave increases as the wave spreads out from the source of the sound. True or false

Physics
1 answer:
ludmilkaskok [199]3 years ago
3 0

Answer:False

Explanation:

the intensity of the sound wave decreases with increasing distance from the source.

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Match each scientist with his major discovery or contribution. 1. Ernest Rutherford discovery of the electron 2. James Chadwick
Shkiper50 [21]

1. Ernerst Rutherford =====> Discovery of Atomic Nucleus

2. James Chadwick =====> Discovery of Neutron

3. Henri Becquerel =====> Discovery of Radioactivity

4. J.J. Thomson =====> Discovery of Electron

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3 years ago
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Determine the internal resistance of the capacitor.
11Alexandr11 [23.1K]

Answer:

The equivalent series resistance of a capacitor is the internal resistance that appears in series with the capacitance of the device. Almost all capacitors exhibit this property at varying degrees depending on the construction, dielectric materials, quality, and reliability of the capacitor.

Explanation:

Hope this helps

7 0
2 years ago
A 0.5 kg block is attached to a spring (k = 12.5 N/m). The damped frequency is 0.2% lower than the natural frequency, (a) What i
Tju [1.3M]

Answer:

a)  1.58 kg s^{-1}

b)  x_m e^{-1.58t}   x_m is initial amplitude

c) 5 kg s^{-1}

Explanation:

given data:

mass =0.5 kg

k = 12.5 N/m

from the data given

a) w_d = w_o - \frac{0.2}{100}w_o

= w_o - 0.002w_o = 0.99w_o

w_d = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2}

0.998w_o = \sqrt{w_o^2 - \frac{b^2}{4m^2}

(0.998w_o)^2 = w_o^2 -\frac{b^2}{1}

b^2 = w_o^2 -(0.998w_o)^2

b^2 = w_o^2(1-0.998^2) = 3.996 *10^{-3} w_o^2

b = w_o\sqrt{3.996*10^{-3}}

b = \frac{12.5}{0.5}\sqrt{3.996*10^{-3}} = 1.58 kg s^{-1}

b)x = x_m e^{\frac{-bt}{2m}}

   x = x_m e^{-1.58t}{1} = x_m e^{-1.58t}    where x_m is initial amplitude

c) critical damping amplitude

c_c =2\sqrt{km} = 2\sqrt{12.5*.5} = 5 kg s^{-1}

5 0
3 years ago
Which answer shows a balanced nuclear equation for the alpha decay of plutonium-239
Sergio [31]

Answer: An alpha particle is a helium-4 nucleus, it contains gwo protons and two neutrons. So, alpha decay, giving off an alpha particle, does the following:

1)reduces the atomic number (number of just protons) by two. Plutonium has atomic number of 94, take two protons away and you are left with 92 which corresponds to uranium.

2) reduces the mass number (protons plus neutrons) by 4. If you start with 239 and subtract two protons and two neutrons, you are left with 235.

Explanation:

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Answer:

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