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Novosadov [1.4K]
3 years ago
13

Two pianos each sound the same note simultaneously, but they are both out of tune. On a day when the speed of sound is 349 m/s,

piano A produces a wavelength of 0.766 m, while piano B produces a wavelength of 0.776 m. How much time separates successive beats?
Physics
1 answer:
SSSSS [86.1K]3 years ago
4 0

Answer:

Time period between the successive beats will be 0.1703 sec

Explanation:

We have given speed of the sound v = 349 m/sec

Wavelength of piano A\lambda _A=0.766m

Wavelength of piano  B\lambda _B=0.776m

So frequency of piano A f_1=\frac{v}{\lambda _1}=\frac{349}{0.766}=455.61Hz

Frequency of piano B f_2=\frac{v}{\lambda _1}=\frac{349}{0.776}=449.74Hz

So beat frequency f = 455.61 - 449.74 = 5.87 Hz

So time period T=\frac{1}{f}=\frac{1}{5.87}=0.1703sec

So time period between the successive beats will be 0.1703 sec

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Order the speed of sound through these materials from the slowest to the fastest.
Sholpan [36]

Speed of sound in cold air < Speed of sound in Warm air < Speed of sound in hot molten lead < Speed of sound in water

Explanation:

Step 1:

Speed of sound in water varies from 1450 to 1498 meters per second

Speed of sound in Hot Molten lead is approximately 1210 meters per second

Speed of sound in warm air is approximately 338.89 meters per second

Speed of sound in cold air is approximately 293.33 meters per second

Step 2:

In warm air sound travels faster than that of sound travelling nature in cold air.

∴ Speed of sound in cold air < Speed of sound in Warm air < Speed of sound in hot molten lead < Speed of sound in water

Speed of sound in cold air the slowest while Speed of sound in water is the fastest mean.

8 0
2 years ago
How much money would be saved by turning off one 100.0-W lightbulb 3.0 h/day for 365 days if the
Pavlova-9 [17]

Answer:

the money that would be saved is $13.14.

Explanation:

Given;

power consumed by the light bulb, P = 100 W = 0.1 kW

time of running the bulb, t = 3 hours for 365 days = 1,095 hours

cost rate of power consumption, C = $0.12 per kWh

Energy consumed by the light bulb for the given days;

E = Pt

E = 0.1 kW  x 1,095 hr

E = 109.5 kWh

Cost of energy consumed = 109.5 kWh   x   $0.12 / kWh

                                            = $13.14

Therefore, the money that would be saved is $13.14.

3 0
2 years ago
I need so much help 50 points!!!!!!<br><br>complete the graph
motikmotik

<u>First Symbol </u>: Cobalt (Co)

Its Group Number - 9

Its Period Number - 4

Its Family Name - Transition Metal

<u>Second Symbol</u> : Silicon (Si)

Its Group Number - 14

Its Period Number - 2

Its Family Name - Semiconductor

<u>Third Symbol</u> : Astatine (At)

Its Group Number - 17

Its Period Number - 6

Its Family Name - Halogen

<u>Fourth Symbol </u>: Magnesium (Mg)

Its Group Number - 2

Its Period Number - 3

Its Family Name - Alkaline Earth Metal

<u>Fifth Symbol</u> : Xenon (Xe)

Its Group Number - 18

Its Period Number - 5

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6 0
3 years ago
A rabbit runs 4.4 m across a lawn, stops, then runs 2.2 m back in the opposite direction. What is the rabbit’s displacement from
Alla [95]

Answer:

it is 2.2 m

Explanation:

because he goes back 2.2 m so 4.4 minus 2.2 equals 2.2

6 0
3 years ago
A 41-turn square coil of area 0.074 m2 and a 123-turn circular coil are both placed perpendicular to the same changing magnetic
vesna_86 [32]

Answer:

<h3>The area of second coil is ≅ 0.025 m^{2}</h3>

Explanation:

Given :

No. of turns in the first coil N_{1} = 41

No. of turns in the second coil N_{2}  = 123

Area of first coil A_{1} = 0.074 m^{2}

According to the law of electromagnetic induction,

Induced emf = -N \frac{d \phi}{dt}

Where \phi = magnetic flux.

Since given in question emf of both coil is same so we compare above equation.

    -\frac{N_{1} d\phi _{1}   }{dt_{1} }  = -\frac{N_{2} d\phi _{2}   }{dt_{2} }

   \frac{N_{1} A_{1}   dB_{1}  }{dt_{1} }  = \frac{N_{2} A_{2} dB_{2}     }{dt_{2} }

        A_{2} = \frac{N_{1} A_{1}  }{N _{2}  }

        A_{2} = \frac{41 \times 0.074 }{123  }

        A_{2} = 0.0246 = 0.025 m^{2}

Therefore, the area of second coil is ≅ 0.025 m^{2}

4 0
3 years ago
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