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natulia [17]
3 years ago
10

What weather could create the best surfing waves?

Physics
2 answers:
charle [14.2K]3 years ago
8 0
C because if you go to the beach you see the huge waves and normaly its windy if not then wind is coming from another direction
Alex Ar [27]3 years ago
5 0

Answer:

Correct answer is C ( Strong winds that have been blowing for a long time with a long fetch)

Explanation:

The size of a wave for surfing depends on the speed of wind it means if the speed of wind will larger(Strong wind) then it will form a larger wave.  Best surfing wave also depends on duration for which wind blows, if wind blow for long time then it will create larger wave which is suitable for surfing  . waves ideal for surfing also depends on fetch it means if wind affect the larger area then a larger wave will be created which is suitable for surfing.

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How can an object have high speed, but a low (or zero) velocity.
Ne4ueva [31]
Something that goes round in circles could have a high speed but zero average velocity.
6 0
3 years ago
Physical science
attashe74 [19]

Answer:

0.56 km/s

Explanation:

We will define a single system of units for measurement, for this case meters per second [m/s]. That is, we must convert the rest of units such as centimeters per second and kilometers per second to meters per second.

560[\frac{cm}{s}]*(\frac{1m}{100cm} )=5.6[m/s]\\0.56[\frac{km}{s}]*(\frac{1000m}{1km} )=560[m/s]

Therefore the speed of 0.56 [km/s] is the greatest of all

3 0
4 years ago
juan aims his boat directly across a river flowing at 8mi/hr. juan's boat travels at 6 miles/hr in still water. how was does jua
son4ous [18]
6mph:

Suppose the boat is traveling on a y axis. The river flow acts on the x axis. Motion on each axes are independent. The linear speed of the boat is not changed. Furthermore the projectile motion is changing, but you're specifically asking about the linear speed of the boat which is unchanged.
4 0
3 years ago
A radio transfers 270 j of energy electrically, 94.5 j is transferred by sound. Work out the efficiency of this radio. Give your
MakcuM [25]

Answer:

35%

Explanation:

Given data

Amount of energy transferred (Input) = 270J

Amount of energy converted to sound (Output)=  94.5J

Efficiency = output/input*100

Efficiency= 94.5/270*100

Efficiency=0.35*100

Efficiency=35%

Hence the efficiency is 35%

6 0
3 years ago
The intensity of the sound from a certain source is measured at two points along a line from the source. The points are separate
Artemon [7]

Answer:

The source is at a distance of 4.56 m from the first point.

Solution:

As per the question:

Separation distance between the points, d = 11.0 m

Sound level at the first point, L = 66.40 dB

Sound level at the second point, L'= 55.74 dB

Now,

L = 10log_{10}\frac{I}{I_{o}}          

I = I_{o}10^{\frac{L}{10}} = I_{o}10^{0.1L} = 10^{- 12}\times 10^{0.1\times 66.40} = 10^{- 5.36}      

L' = 10log_{10}\frac{I'}{I_{o}}

I' = I_{o}10^{\frac{L'}{10}} = 10^{- 12}\times 10^{0.1\times 55.74} = 10^{- 6.426}        

where

I_{o} = 10^{- 12} W/m^{2}

I = Intensity of sound

Now,

I = \frac{P}{4\pi R^{2}}

Similarly,

I' = \frac{P}{4\pi (R + 11.0)^{2}}

Now,

\frac{I}{I'} = \frac{(R + 11.0)^{2}}{R^{2}}

\frac{10^{- 5.36}}{10^{- 6.426}} = \frac{(R + 11.0)^{2}}{R^{2}}

R ^{2} + 22R + 121 = 11.64R^{2}}

10.64R ^{2} - 22R - 121 = 0

Solving the above quadratic eqn, we get:

R = 4.56 m

8 0
3 years ago
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