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Ymorist [56]
2 years ago
11

Two swift canaries fly toward each other, each moving at 15.0 m/s relative to the ground, each warbling a note of frequency 1750

Hz. (a) What frequency note does each bird hear from the other one? (b) What wavelength will each canary measure for the note from the other one?
Physics
1 answer:
lesya692 [45]2 years ago
3 0

Answer:

Part a)

f = 1911.5 Hz

Part b)

\lambda = 0.186 m

Explanation:

Here the source and observer both are moving towards each other

so we know that the apparent frequency is given as

f' = f_0 (\frac{v + v_o}{v - v_s})

here we know that

f_0 = 1750 Hz

v_o = 15 m/s

v_s = 15 m/s

now we will have

f' = (1750)(\frac{340 + 15}{340 - 15})

f' = 1911.5 Hz

Part b)

Apparent wavelength is given by the formula

\lambda = \frac{v_{relative}}{f_{app}}

here we will have

\lambda = \frac{340 + 15}{1911.5}

\lambda = 0.186 m

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A microprocessor scans the status of an output I/O device every 20 ms. This is accomplished by means of a timer alerting the pro
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0.0000045 s

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\dfrac{1}{f}=\dfrac{1}{8\times 10^6}=1.25\times 10^{-7}\ s

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12\times 1.25\times 10^{-7}=0.0000015\ s

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3\times 0.0000015=0.0000045\ s

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1 year ago
A person wants to lose weight by "pumping iron". The person lifts an 80 kg weight 1 meter. How many times must this weight be li
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Answer:

37357 sec  

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Explanation:

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lifts required = \frac{(14.644 x 10^6)}{784}

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Your friend Amanda suffers from a condition that reduces her blood's ability to carry oxygen.which of the following is the name
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2 years ago
Given three capacitors, c1 = 2.0 μf, c2 = 1.5 μf, and c3 = 3.0 μf, what arrangement of parallel and series connections with a 12
Lesechka [4]

Answer:

Connect C₁ to C₃ in parallel; then connect C₂ to C₁ and C₂ in series. The voltage drop across C₁ the 2.0-μF capacitor will be approximately 2.76 volts.

-1.5\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-3.0\;\mu\text{F}-\end{array}]-.

Explanation:

Consider four possible cases.

<h3>Case A: 12.0 V.</h3>

-\begin{array}{c}-{\bf 2.0\;\mu\text{F}-}\\-1.5\;\mu\text{F}- \\-3.0\;\mu\text{F}-\end{array}-

In case all three capacitors are connected in parallel, the 2.0\;\mu\text{F} capacitor will be connected directed to the battery. The voltage drop will be at its maximum: 12 volts.

<h3>Case B: 5.54 V.</h3>

-3.0\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-1.5\;\mu\text{F}-\end{array}]-

In case the 2.0\;\mu\text{F} capacitor is connected in parallel with the 1.5\;\mu\text{F} capacitor, and the two capacitors in parallel is connected to the 3.0\;\mu\text{F} capacitor in series.

The effective capacitance of two capacitors in parallel is the sum of their capacitance: 2.0 + 1.5 = 3.5 μF.

The reciprocal of the effective capacitance of two capacitors in series is the sum of the reciprocals of the capacitances. In other words, for the three capacitors combined,

\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_3}+ \dfrac{1}{C_1+C_2}} = \frac{1}{\dfrac{1}{3.0}+\dfrac{1}{2.0+1.5}} = 1.62\;\mu\text{F}.

What will be the voltage across the 2.0 μF capacitor?

The charge stored in two capacitors in series is the same as the charge in each capacitor.

Q = C(\text{Effective}) \cdot V = 1.62\;\mu\text{F}\times 12\;\text{V} = 19.4\;\mu\text{C}.

Voltage is the same across two capacitors in parallel.As a result,

\displaystyle V_1 = V_2 = \frac{Q}{C_1+C_2} = \frac{19.4\;\mu\text{C}}{3.5\;\mu\text{F}} = 5.54\;\text{V}.

<h3>Case C: 2.76 V.</h3>

-1.5\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-3.0\;\mu\text{F}-\end{array}]-.

Similarly,

  • the effective capacitance of the two capacitors in parallel is 5.0 μF;
  • the effective capacitance of the three capacitors, combined: \displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_2}+ \dfrac{1}{C_1+C_3}} = \frac{1}{\dfrac{1}{1.5}+\dfrac{1}{2.0+3.0}} = 1.15\;\mu\text{F}.

Charge stored:

Q = C(\text{Effective}) \cdot V = 1.15\;\mu\text{F}\times 12\;\text{V} = 13.8\;\mu\text{C}.

Voltage:

\displaystyle V_1 = V_3 = \frac{Q}{C_1+C_3} = \frac{13.8\;\mu\text{C}}{5.0\;\mu\text{F}} = 2.76\;\text{V}.

<h3 /><h3>Case D: 4.00 V</h3>

-2.0\;\mu\text{F}-1.5\;\mu\text{F}-3.0\;\mu\text{F}-.

Connect all three capacitors in series.

\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_1} + \dfrac{1}{C_2}+\dfrac{1}{C_3}} =\frac{1}{\dfrac{1}{2.0} + \dfrac{1}{1.5}+\dfrac{1}{3.0}} =0.667\;\mu\text{F}.

For each of the three capacitors:

Q = C(\text{Effective})\cdot V = 0.667\;\mu\text{F} \times 12\;\text{V} = 8.00\;\mu\text{C}.

For the 2.0\;\mu\text{F} capacitor:

\displaystyle V_1=\frac{Q}{C_1} = \frac{8.00\;\mu\text{C}}{2.0\;\mu\text{F}} = 4.0\;\text{V}.

6 0
1 year ago
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