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3 years ago

What is most likely source of the waves approaching this coastline

1 answer:

hjlf3 years ago

4 0

Assuming that you mean waves on the surface of a large body of water (like an ocean or lake), the most probable cause of these is the wind. The friction between the moving air and the water surface makes the water move. Hope this answers the question. Have a nice day.

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Two cars travel westward along a straight highway, one at a constant velocity of 85 km/h, and the other at a constant velocity o

spayn [35]

To solve letter a:

d1 = 85t1 = 16 km,
85t1 = 16,
t1 = 16 / 85 = 0.1882 h = 11.29 min.

d2 = 115t2 = 16 km,
115t2 = 16,
t2 = 16 / 115 = 0.139 h = 8.35 min.

t1 - t2 = 11.29 - 8.35 = 2.94 min.
Car #2 arrives 2.94 minutes sooner.

To solve letter b:

15 min = 1/4 h = 0.25 h.
d1 = d2,
115t = 85(t + 0.25),
115t = 85t + 21.25,
115t - 85t = 21.25,
30t = 21.25,
t = 21.25 / 30 = 0.71 h,

d = 115 * 0.71 = 81.65 km.

8 0

3 years ago

At sunset, the sun appears reddish. What is MOST LIKELY the reason for this phenomenon?

Alexeev081 [22]

The only correct statement on the list is choice-A./

3 0

4 years ago

Read 2 more answers

The time it takes two successive crests of an ocean wave to pass a given point is called a _____.

nika2105 [10]

The answer is Period

3 0

3 years ago

Read 2 more answers

The lower the value of the coefficient of friction,the_the resistance to sliding

malfutka [58]

Answer:

lower

Explanation:

The lower the value of the coefficient of friction, the lower the resistance to sliding.

The coefficient of friction is the ratio of the frictional force and the normal force pressing two surfaces in contact together.

U = $\frac{F}{N}$

U is the coefficient of friction

F is the frictional force

N is the normal force

We see that coefficient of friction is directly proportional to frictional force.

7 0

3 years ago

Calculate the wavelength of each frequency of electromagnetic radiation: a. 100.2 MHz (typical frequency for FM radio broadcasti

Natalka [10]

Answer:

a). 100.2 MHz (typical frequency for FM radio broadcasting)

The wavelength of a frequency of 100.2 Mhz is 2.99m.

b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)

The wavelength of a frequency of 1070 khz is 280.3 m.

c. 835.6 MHz (common frequency used for cell phone communication)

The wavelength of a frequency of 835.6 Mhz is 0.35m.

Explanation:

The wavelength can be determined by the following equation:

$c = \lambda \cdot \nu$ (1)

Where c is the speed of light, $\lambda$ is the wavelength and $\nu$ is the frequency.

Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave.

a). 100.2 MHz (typical frequency for FM radio broadcasting)

Then, $\lambda$ can be isolated from equation 1:

$\lambda = \frac{c}{\nu}$ (2)

since the value of c is $3x10^{8}m/s$ . It is necessary to express the frequency in units of hertz.

$\nu = 100.2 MHz . \frac{1x10^{6}Hz}{1MHz}$ ⇒ $100200000Hz$

But $1Hz = s^{-1}$

$\nu = 100200000s^{-1}$

Finally, equation 2 can be used:

$\lambda = \frac{3x10^{8}m/s}{100200000s^{-1}}$

$\lambda = 2.99 m$

Hence, the wavelength of a frequency of 100.2 Mhz is 2.99m.

b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)

$\nu = 1070kHz . \frac{1000Hz}{1kHz}$ ⇒ $1070000Hz$

But $1Hz = s^{-1}$

$\nu = 1070000s^{-1}$

Finally, equation 2 can be used:

$\lambda = \frac{3x10^{8}m/s}{1070000s^{-1}}$

$\lambda = 280.3 m$

Hence, the wavelength of a frequency of 1070 khz is 280.3 m.

c. 835.6 MHz (common frequency used for cell phone communication)

$\nu = 835.6MHz . \frac{1x10^{6}Hz}{1MHz}$ ⇒ $835600000Hz$

But $1Hz = s^{-1}$

$\nu = 835600000s^{-1}$

Finally, equation 2 can be used:

$\lambda = \frac{3x10^{8}m/s}{835600000s^{-1}}$

$\lambda = 0.35 m$

Hence, the wavelength of a frequency of 835.6 Mhz is 0.35m.

6 0

3 years ago