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aniked [119]
3 years ago
14

after singing for a few minutes the soloist rejoins the choir a second choir, consist of 90 additional singers then joins in eac

h group of 10 singers sings with the same intensity as the first 10 person choir now how many decibels of sound intensity as the first 10 person choir now how many decibels of sound intensity are being produced
Physics
1 answer:
vredina [299]3 years ago
7 0
If the soloist produces "x" decibels and the 10-person choir produces "y" decibels, combined they will produce "x+y" decibels. 

The second choir has 90 additional singers, we base our description on the first choir. If a 10-person choir produces "x+y" decibels, then the 90 person choir produces 10 (x+y) decibels. 
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What are two types of electric currents?
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Answer:

There are two different types of current in widespread use today. They are direct current, abbreviated DC, and alternating current, abbreviated AC. In a direct current, the electrons flow in one direction.

6 0
3 years ago
Fe has the electron configuration [Ar]3d64s2. What is the magnitude of the orbital angular momentum for its most energetic elect
salantis [7]

Answer : The magnitude of the orbital angular momentum for its most energetic electron is, \sqrt{6}\hbar

Explanation :

The formula used for orbital angular momentum is:

L=\sqrt{l(l+1)}\hbar

where,

L = orbital angular momentum

l = Azimuthal quantum number

As we are given the electronic configuration of Fe is, [Ar]3d^64s^2

Its most energetic electron will be for 3d electrons.

The value of azimuthal quantum number(l) of d orbital is, 2

That means, l = 2

Now put all the given values in the above formula, we get:

L=\sqrt{2(2+1)}\hbar

L=\sqrt{6}\hbar

Therefore, the magnitude of the orbital angular momentum for its most energetic electron is, \sqrt{6}\hbar

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3 years ago
A metal bar magnet produces a magnetic field in space surrounding it. The point
MatroZZZ [7]
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3 0
3 years ago
A 0.150-kg glider is moving to the right with a speed of 0.80 m/s on a frictionless, horizontal air track. The glider has a head
victus00 [196]

Answer:

The final speed of 0.150-kg glider after collision is 3.2 m/s to the left

The final speed of 0.300-kg glider after collision is 0.20 m/s to the left

Explanation:

Given;

mass of glider moving to the right, m₁ = 0.150-kg

mass of glider moving to the left, m₂ = 0.300-kg

initial speed of glider moving to the right, u₁ = 0.80 m/s

initial speed of glider moving to the left, u₂ = 2.20 m/s

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.15 x 0.8) + (0.3 x - 2.2) = 0.15v₁ + 0.3v₂

-0.54 = 0.15v₁ + 0.3v₂

0.15v₁ + 0.3v₂ = - 0.54 -----------equation (i)

Again, their relative velocity after collision is given as;

u₁ - u₂ = v₂ - v₁

0.8 - (-2.2) = v₂ - v₁

3 = v₂ - v₁

v₂ =  v₁ + 3  ------------equation (ii)

Substitute v₂ in equation (i)

0.15v₁ + 0.3v₂ = - 0.54

0.15v₁ + 0.3(v₁ + 3 ) = - 0.54

0.15v₁ + 0.3v₁ + 0.9 = - 0.54

0.45v₁  = - 0.54 - 0.9  

0.45v₁  = -1.44

v₁ = -1.44 /0.45

v₁ = - 3.2 m/s

Thus, the final speed of 0.150-kg glider after collision is 3.2 m/s to the left

From equation (ii), v₂ = v₁ + 3

v₂ = -3.2 + 3

v₂ = - 0.20 m/s

Therefore, the final speed of 0.300-kg glider after collision is 0.20 m/s to the left

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