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Korvikt [17]
3 years ago
8

By what factor must the amplitude of a sound wave be increased in order to increase the intensity by a factor of 9?a. 9 b. 2 c.

4 d. 3
Physics
2 answers:
jolli1 [7]3 years ago
6 0

Answer:

3 times option (d) is correct

Explanation:

Initial intensity = Io

final intensity = 9 Io

initial amplitude = A

Final amplitude = ?

let the final amplitude is A'.

Intensity of a wave is directly proportional to the square of the amplitude of the wave.

I ∝ A²

So, Io ∝ A²   ... (1)

9Io ∝ A'²     .... (2)

Divide (2) by (1)

9 = A'²/A²

A' = 3 A

So, the amplitude becomes 3 times.

Marta_Voda [28]3 years ago
4 0

Answer:

option D

Explanation:

given,

intensity\ \alpha \ (Amplitude)^2

increase the intensity by factor of 9

    I₁ = I₀

    I₂ = 9 I₀

now,

\dfrac{I_1}{I_2}=\dfrac{A_1^2}{A_2^2}

\dfrac{I_0}{9I_0}=(\dfrac{A_1}{A_2})^2

(\dfrac{A_1}{A_2})^2=\dfrac{1}{9}

\dfrac{A_1}{A_2}=\dfrac{1}{3}

      A₂ = 3 A₁

hence, amplitude increase with the factor of 3

so, the correct answer is option D

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ω = 2π/60
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v = 8 x 10⁻³ m/s
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If the magnitude of the resultant force is to be 500n, directed along the positive y axis, determine the magnitude of force f an
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The effective nuclear charge experienced by the outermost electron of Na is different than the effective nuclear charge experien
Nesterboy [21]

Answer:

B) Na has a lower first ionization energy than Ne.  

Explanation:

The atomic number¹ for Na has a value of 11 while in the case of Ne this value is 10. That means that Sodium (Na) has a total number of 11 protons, 11 neutrons and 11 electrons (since it is electrically neutral²). For the case of Neon (Ne) it has 10 protons, 10 neutrons and 10 electrons.

As the atomic number increases, the atomic radius³ shrinks (the orbitals are closer to the nucleus) as a consequence of the electric force. For the case of sodium (Na) the electron in the outermost orbital will experience a lower electric force than the electron placed in the outermost orbital in the atom of Neon (Ne).

Although, the sodium’s atom has more protons and therefore electrons, these eleven electrons will be organized according with the electronic configuration⁴ in the different shells (orbitals) of probabilities of their positions around the atom.

The electronic configuration for Na is:

1s²2s²2p⁶3s¹

The electronic configuration for Ne is:

1s²2s²2p⁶

Since Na needs another orbital to placed its outermost electron, the atomic radius will have a greater value than Ne. The electric force is inversely proportional to the square of the distance between two charged particles, as is established in Coulomb’s law:

F = \kappa_{0} \frac{q1q2}{r^{2}}    (1)

Where q1 and q2 are the charges, \kappa_{0} is the proportionality constant and r is the distance between the two charges.

Hence, the electron in the outermost orbital of Ne is submitted to a greater electric force according with equation 1, the required energy to remove it (ionization energy⁵) will be greater than in the case of Na (<u>for that case will be the first ionization energy</u>).

¹Atomic number: The number of protons or electrons in an atom.

²Electricaly neutral: All the charges are balanced (same number of positive charges and negative charges).

³Atomic radius: Distance between the center of the nucleus and an electron placed in the outermost orbital for a specific atom.

⁴Electronic configuration: Show how the electrons of an atom will be arranged in different orbitals according with the fact that each orbital has a specific number of electrons that can be held.

⁵Ionization energy: Energy required to remove an electron from an atom.

Key values:

First ionization energy of Na: 495 kJ/mol

First ionization energy of Ne: 2080 kJ/mol

Atomic radius of Na: 2.27 Å

Atomic radius of Ne: 1.54 Å

Atomic number of Na: 11

Atomic number of Ne: 10

3 0
3 years ago
Read 2 more answers
A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?
KATRIN_1 [288]

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU.  What is its orbital period?

Answer:

The value  is  T_R  = 11.8 \  days  

Explanation:

From the question we are told that

   The semi - major axis of the rocky debris  a_R = 45.0\  AU

   The semi - major axis of  Planet D is  a_D  = 60 \  AU

    The orbital  period of planet D is  T_D = 18.164 \  days

Generally from Kepler third law

          T \  \ \alpha \ \ a^{\frac{3}{2} }

Here T is the  orbital period  while a is the semi major axis

So  

        \frac{T_D}{T_R}  =  \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}

=>     T_R  = T_D *  [\frac{a_R}{a_D} ]^{\frac{3}{2} }  

=>     T_R  = 18.164  *  [\frac{ 45}{60} ]^{\frac{3}{2} }

=>      T_R  = 11.8 \  days  

   

7 0
3 years ago
A 81 kg person sits on a 3.8 kg chair. Each leg of the chair makes contact with the floor in a circle that is 1.2 cm in diameter
tankabanditka [31]

Answer:

1.9 MPa

Explanation:

Mass of person = 81 kg

Mass of chair = 3.8 kg

Diameter of contact point = 1.2 cm = D

Radius of contact point = 1.2/2 = 0.6 cm

Total mass of chair and person = 81 + 3.8 = 84.8 kg = M

Acceleration due to gravity = 9.81 m/s²

Force acting on the floor

<em>F = Mg</em>

<em>⇒F = 84.8×9.81</em>

<em>⇒F = 831.888 N</em>

Area of the contact point

<em>A = πR²</em>

<em>⇒A = π0.006²</em>

<em>⇒A = π0.000036 m²</em>

Area of the four points is

<em>4A = 0.000144π m²</em>

Pressure

p=\frac{F}{A}\\\Rightarrow p=\frac{831.888}{0.000144\pi}\\\Rightarrow p=1838876.21\ Pa=1.83887621\times 10^6\ Pa=1.9 MPa

Pressure exerted on the floor by each leg of the chair is 1.9 MPa

5 0
3 years ago
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