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shepuryov [24]
3 years ago
13

Witch is the fastest way to send information, useing sound waves, light waves, or water waves

Physics
1 answer:
malfutka [58]3 years ago
4 0
The answer is light waves
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7. If 8 million kg of water flows over Niagara Falls each second, calculate the power available at the bottom of the falls.
Alexxx [7]

Answer:

The power will be "3.92×10⁹ Watts". A further explanation is given below.

Explanation:

The given values as per the question,

Rate,

= 8 million kg

Distance,

= 50 m

Gravity,

= 9.8 m/s²

As we know,

The power will be:

⇒ Power = Rate\times Distance\times  Gravity

On putting the values, we get

⇒             =  8\times 10^6\times 50\times 9.8

⇒             =3.92\times 10^9 \  Watts

7 0
2 years ago
Radio waves just like light waves can be reflected refracted and diffracted and polarized.
hoa [83]
<span>Radio waves just like light waves can be reflected refracted and diffracted and polarized.  The answer is True. </span>These characteristics are the common phenomena for electromagnetic (EM)  waves, and Radio Waves are electromagnetic Waves so much so that they obey reflection, refraction, and diffraction. 
5 0
3 years ago
Read 2 more answers
In which test did the air parcel rise the highest? Is there a pattern in the relationship between starting air temperature and p
wel

Answer:

xitxitx

Explanation:

ufxigxigxitcutcocyocoycitcotcitcohcohxitxigcigxigxkgxigxigxigxigxigxigxitx8y

5 0
3 years ago
When two substances are mixed together, a color change is observed. Does this mean that a chemical reaction has definitely occur
Elanso [62]
Nope, color change can also occur during a physical change. 
7 0
3 years ago
Two identical strings, of identical lengths of 2.00 m and linear mass density of μ=0.0065kg/m, are fixed on both ends. String A
kolezko [41]

Answer:

beat frequency = 13.87 Hz

Explanation:

given data

lengths l = 2.00 m

linear mass density μ = 0.0065 kg/m

String A is under a tension T1 = 120.00 N

String B is under a tension T2 = 130.00 N

n = 10 mode

to find out

beat frequency

solution

we know here that length L is

L = n × \frac{ \lambda }{2}      ........1

so  λ = \frac{2L}{10}  

and velocity is express as

V = \sqrt{\frac{T}{\mu } }    .................2

so

frequency for string A = f1 = \frac{V1}{\lambda}

f1 = \frac{\sqrt{\frac{T}{\mu } }}{\frac{2L}{10}}

f1 = \frac{10}{2L} \sqrt{\frac{T1}{\mu } }      

and

f2 = \frac{10}{2L} \sqrt{\frac{T2}{\mu } }

so

beat frequency is = f2 - f1

put here value

beat frequency = \frac{10}{2*2} \sqrt{\frac{130}{0.0065}}  - \frac{10}{2*2} \sqrt{\frac{120}{0.0065} }

beat frequency = 13.87 Hz

6 0
3 years ago
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