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zloy xaker [14]

3 years ago

8

We know the frequency range of certain sounds are: 400-560 Hz, what are the ranges of wavelength in meters when the signal trans

mits in air?

1 answer:

Ksivusya [100]3 years ago

8 0

Answer:

Range of wavelength will be $5.35\times 10^5m$ to $7.5\times 10^5m$

Explanation:

We have given range of frequency is 400-560 Hz

Speed of the light $c=3\times 10^8m/sec$

We have to find the range of the wavelength of signal transmitted

Ween know that velocity is given by $v=\lambda f$ , here $\lambda$ is wavelength and f is frequency

So for 400 Hz frequency wavelength will be $\lambda =\frac{c}{f}=\frac{3\times 10^8}{400}=7.5\times 10^5m$

And wavelength for frequency 560 Hz $\lambda =\frac{c}{f}=\frac{3\times 10^8}{560}=5.35\times 10^5m$

So range of wavelength will be $5.35\times 10^5m$ to $7.5\times 10^5m$

You might be interested in

A carpenter is driving a 15.0-g steel nail into a board. His 1.00-kg hammer is moving at 8.50 m/s when it strikes the nail. Half

bekas [8.4K]

Answer: The increase in temperature of the nail after the three blows is 8.0636 Kelvins. The correct option is (d).

Explanation:

Kinetic energy of the hammer ,K.E.=

$\frac{1}{2}mv^2=\frac{1}{2}1.00 kg\times (8.50 m/s)^2=36.125 J$

Half of the kinetic energy of the hammer is transformed into heat in the nail.

Energy transferred to the nail in one blow =

$\frac{1}{2}K.E.=\frac{1}{2}\times 36.125 J=18.0625 J$

Total energy transferred after 3 blows,Q = $3\times 18.0625 J=54.1875 J$

Mass of the nail = 15 g = 0.015 kg

Change in temperature = $\Delta T$

Specif heat of the steel = c = 448 J/kg K

$Q=mc\Delta T$

$54.1875 J=0.015 kg\times 448 J/kg K\times \Delta T$

$\Delta T=8.0636 K\approx 8.1 K$

The increase in temperature of the nail after the three blows is 8.1 Kelvins.Hence, correct option is (d).

4 0

3 years ago

A car battery with a 12 V emf and an internal resistance of 0.11 Ω is being charged with a current of 56 A. What are (a) the pot

denpristay [2]

Answer:

Part a)

V = 18.16 V

Part b)

$P_r = 345 Watt$

Part c)

P = 672 Watt

Part d)

V = 5.84 V

Part e)

$P_r = 345 Watt$

Explanation:

Part a)

When battery is in charging mode

then the potential difference at the terminal of the cell is more than its EMF and it is given as

$\Delta V = E + i r$

here we have

$E = 12 V$

$i = 56 A$

$r = 0.11$

now we have

$\Delta V = 12 + (0.11)(56) = 18.16 V$

Part b)

Rate of energy dissipation inside the battery is the energy across internal resistance

so it is given as

$P_r = i^2 r$

$P_r = 56^2 (0.11)$

$P_r = 345 W$

Part c)

Rate of energy conversion into EMF is given as

$P_{emf} = i E$

$P_{emf} = (56)(12)$

$P_{emf} = 672 Watt$

Now battery is giving current to other circuit so now it is discharging

now we have

Part d)

$V = E - i r$

$V = 12 - (56)(0.11)$

$V = 12 - 6.16 = 5.84 V$

Part e)

now the rate of energy dissipation is given as

$P_r = i^2 r$

$P_r = 56^2 (0.11)$

$P_r = 345 W$

7 0

3 years ago

When a 2.50 kg object is hung vertically on a certain light spring described by hookes law the spring stretches 2.76 cm.

Readme [11.4K]

F=K*X,
F=M*a

M*a=K*X

2.5*9.81=K*0.0276

24.525=K*0.0276

24.525/0.0276=K

K= 888.6 N/m ---- force constant

assuming 2.5 refers to the new extension, just divide F/ 0.025
to get

981N/m

8 0

4 years ago

What is one common product that uses microwaves

AVprozaik [17]

Electromagnet Radiation

7 0

3 years ago

Consider two ideal gases, A & B, at the same temperature. The rms speed of the molecules of gas A is twice that of gas B. Ho

uysha [10]

Answer:

option (d)

Explanation:

The relation between the rms velocity and the molecular mass is given by

v proportional to \frac{1}{\sqrt{M}} keeping the temperature constant

So for two gases

$\frac{v_{A}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}$

$\frac{2v_{B}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}$

${\frac{M_{B}}{M_{A}}} = 4$

${\frac{M_{B}}{4}} = M_{A}$

7 0

4 years ago