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Softa [21]
3 years ago
8

This problem has been solved!See the answerCommunication 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:
mash [69]3 years ago
8 0

Answer:

4000 km

Explanation:

as we know velocity of electromagnetic wave is c

c = 3 * 10^8 m/s

frequency  given (f) = 76 Hz

wavelength ?

using

c =fλ

λ =   \frac{c}{f}

λ =  \frac{3 * 10^8}{76}  ≈4000 km

 so final answer λ = 4000km

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A beaker weighs 0.4N when empty and1.4N when filled with water what does ot weigh when filled with brine of density 1.2 g/cm3
PtichkaEL [24]

Answer:2.47

Explanation:

So, the beaker weighs 1.40N when filled with water, brine of density weighs about 1.7N, you add the density + water. Have a good day!

7 0
3 years ago
Two loudspeakers are 1.60 m apart. A person stands 3.00 m from one speaker and 3.50 m from the other. (a) What is the lowest fre
VMariaS [17]

Answer:

Explanation:

Given

Distance between two loud speakers d=1.6\ m

Distance of person from one speaker x_1=3\ m

Distance of person from second speaker x_2=3.5\ m

Path difference between the waves is given by

x_2-x_1=(2m+1)\cdot \frac{\lambda }{2}

for destructive interference m=0 I.e.

x_2-x_1=\frac{\lambda }{2}

3.5-3=\frac{\lambda }{2}

\lambda =0.5\times 2

\lambda =1\ m

frequency is given by

f=\frac{v}{\lambda }

where v=velocity\ of\ sound\ (v=343\ m/s)

f=\frac{343}{1}=343\ Hz

For next frequency which will cause destructive interference is

i.e. m=1 and m=2

3.5-3=\frac{2\cdot 1+1}{2}\cdot \lambda

\lambda =\frac{1}{3}\ m

frequency corresponding to this is

f_2=\frac{343}{\frac{1}{3}}=1029\ Hz

for m=2

3.5-3=\frac{5}{2}\cdot \lambda

\lambda =\frac{1}{5}\ m

Frequency corresponding to this wavelength

f_3=\frac{343}{\frac{1}{5}}

f_3=1715\ Hz                        

8 0
4 years ago
Please answer as fast as possible
aleksley [76]
Electric pumps are not useful they clog
6 0
3 years ago
Read 2 more answers
A 6.8 KG object moves with a velocity of 8 M/S what’s it's kinetic energy?
FromTheMoon [43]

Answer:

Explanation:

KE = ½mv² = ½(6.8)8² = 217.6 J

round as appropriate because that result is way too much precision for the inputs provided. Arguably should be 200 J based on the single significant digit of the velocity.

8 0
3 years ago
A steel beam that is 5.50 m long weighs 332 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending
ElenaW [278]

Explanation:

The given data is as follows.

    Length of beam, (L) = 5.50 m

    Weight of the beam, (W_{b}) = 332 N

     Weight of the Suki, (W_{s}) = 505 N

After crossing the left support of the beam by the suki then at some overhang distance the beam starts o tip. And, this is the maximum distance we need to calculate. Therefore, at the left support we will set up the moment and equate it to zero.

                 \sum M_{o} = 0

     -W_{s} \times x + W_{b} \times 1.5 = 0

                x = \frac{W_{b} \times 1.5}{W_{s}}

                   = \frac{332 N \times 1.5}{505 N}

                   = 0.986 m

Hence, the suki can come (2 - 0.986) m = 1.014 from the end before the beam begins to tip.

Thus, we can conclude that suki can come 1.014 m close to the end before the beam begins to tip.

8 0
4 years ago
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