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Vera_Pavlovna [14]
3 years ago
8

Communication with submerged submarines via radio waves is difficult because seawater is conductive and absorbs electromagnetic

waves. Penetration into the ocean is greater at longer wavelengths, so the United States has radio installations that transmit at 76Hz for submarine communications.What is the approximate wavelength of those extremely low-frequency waves?500 km1000 km2000 km4000 km
Physics
1 answer:
REY [17]3 years ago
4 0

Answer:

\lambda=4000\ km

Explanation:

It is given that,

Frequency for submarine communications, f = 76 Hz

We need to find the wavelength of those extremely low-frequency waves. the relation between the wavelength and the frequency is given by :

c=f\times \lambda

\lambda=\dfrac{c}{f}

\lambda=\dfrac{3\times 10^8\ m/s}{76\ Hz}

\lambda=3947368.42\ m

\lambda=3947\ km

or

\lambda=4000\ km

So, the wavelength of those extremely low-frequency waves is 4000 km. Hence, this is the required solution.

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a wrench weighs 5.24 newtons on earth. when it is taken to the Moon, where g =1.16 m/s2 how much does it weigh?
Andrew [12]

“Weight of the wrench” on “the moon” is “6.07 kg”.

<u>Explanation</u>:

Weight of the wrench is 5.24 N  

Weight of the wrench in kilograms = W × g

Taken “g” on the moon is 1.16 \mathrm{m} / \mathrm{s}^{2}

=5.24 \mathrm{N} \times 9.81 \mathrm{m} / \mathrm{s}^{2}=51.352 \mathrm{kg}

Weight of the wrench in kilograms is 51.352 kg.

Formula to calculate weight of the object on the moon is

\frac{\text {weight of the object on earth}}{9.81 \mathrm{m} / \mathrm{s}^{2}} \times 1.16 \mathrm{m} / \mathrm{s}^{2}

Substitute the values given,

=\frac{51.352 \mathrm{kg}}{9.81 \mathrm{m} / \mathrm{s}^{2}} \times 1.16 \mathrm{m} / \mathrm{s}^{2}

=5.234 \times 1.16

= 6.07 kg

Therefore, weight of the wrench on the moon is 6.07 kg.  

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3 years ago
What describes a sound wave as it travels through a medium
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Sound waves in air (and any fluid medium) are longitudinal waves because particles of the medium through which the sound is transported vibrate parallel to the direction that the sound wave moves.
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By what factor does the peak frequency change if the celsius temperature of an object is doubled from 20.0 ∘c to 40.0 ∘c?
mart [117]

Answer:

it increases by a factor 1.07

Explanation:

The peak wavelength of an object is given by Wien's displacement law:

\lambda=\frac{b}{T} (1)

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b is the Wien's displacement constant

T is the temperature (in Kelvins) of the object

given the relationship between frequency and wavelength of an electromagnetic wave:

f=\frac{c}{\lambda}

where c is the speed of light, we can rewrite (1) as

\frac{c}{f}=\frac{b}{T}\\f=\frac{Tc}{b}

So the peak frequency is directly proportional to the temperature in Kelvin.

In this problem, the temperature of the object changes from

T_1 = 20.0^{\circ}+273=293 K

to

T_2 = 40.0^{\circ}+273 = 313 K

so the peak frequency changes by a factor

\frac{f_2}{f_1} \propto \frac{T_2}{T_1}=\frac{313 K}{293 K}=1.07

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Find a glass jar with a screw-top metal lid. Close the lid snugly and put the jar into the refrigerator. Leave it there for abou
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Answer:

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