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-Dominant- [34]
3 years ago
9

A hawk flying at 38 m/s emits a cry whose frequency is 440 Hz. A wren is moving in the same direction as the hawk at 17 m/s. (As

sume the speed of sound is 343 m/s.) (a) What frequency does the wren hear (in Hz) as the hawk approaches the wren? Hz (b) What frequency does the wren hear (in Hz) after the hawk passes the wren?
Physics
1 answer:
Natasha_Volkova [10]3 years ago
3 0

Answer:

frequency wren hear when approaches is 470.29 Hz

frequency wren hear after hawk pass is 415.75 Hz

Explanation:

given data

hawk  velocity Vs= 38 m/s

frequency f = 440 Hz

wren velocity Vo= 17 m/s

speed of sound s = 343 m/s

to find out

What frequency wren hear and What frequency wren hear after hawk pass

solution

we apply here frequency formula that is

f1=f ( \frac{s+V_o}{s+V_s} )    .........1

here f1 is frequency hear by observer

put here all value as Vo and Vs negative because it approaches

f1=f ( \frac{s-V_o}{s-V_s} )  

f1=440 ( \frac{343-17}{343-38} )  

f1 = 470.29 Hz

so frequency wren hear when approaches is 470.29 Hz

and

after passing from equation 1 we take both Vo and Vs as positive

f1=f ( \frac{s+V_o}{s+V_s} )

f1=440 ( \frac{343+17}{343+38} )

f1 = 415.75 Hz

so frequency wren hear after hawk pass is 415.75 Hz

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7 0
3 years ago
If a dog is mass iS 14.3 kg what is it’s weight on earth
BigorU [14]

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Weight of the dog on surface of earth is 140.14 Newton.

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3 0
3 years ago
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>).

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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
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Calculate the standard electrode potential difference (e°) of the daniell cell (at 1 bar) if temperature is 473.15 k.
anzhelika [568]
Missing data in the text of the exercise: The molar concentration of Zinc is 10 times the molar concentration of copper.

Solution:

1) First of all, let's calculate the standard electrode potential difference at standard temperature. This is given by:
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2) To calculate E^0 at any temperature T, we should use Nerst equation:
E^0(T)=E^0- \frac{R T}{z F} \ln  \frac{[Zn]}{[Cu]}
where 
R=8.31 J/(K mol)
T=473.15 K is the temperature in our problem
z=2 is the number of electrons transferred in the cell's reaction
F=9.65\cdot 10^4 C/mol is the Faraday's constant
[Zn] and [Cu] are the molar concentrations of zinc and in copper, and in our problem we have [Zn]=10[Cu].
Using all these data inside the equation, and using E^0=+1.1 V, in the end we find:
E^0(T)=E^0- \frac{R T}{z F} \ln \frac{[Zn]}{[Cu]}=+1.053 V
8 0
3 years ago
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