All categories

2 years ago

The first wave that forms when the wind begins to blow across the ocean surface is a ________.

Physics

1 answer:

seraphim [82]2 years ago

8 0

The first wave that forms when the wind begins to blow across the ocean surface is a <u>capillary wave</u>

The capillary waves represent the initial state of a wave that are produced by vortices in a flat sea. When the winds blow over an ocean, energy is transferred from the wind to the sea. and that produced the waves. They are also presented as ripples.

The capillary waves have a wavelength less than 1.7 cm or 0.68 inches and its speed of 0.2 to 0.3 meter per second.

<h3>What is a capillary wave?</h3>

It is a wave that is travelling along the interface boundary of a fluid and it is influenced by the effect of surface tension and the gravity.

Learn more about capillary wave at brainly.com/question/15663649

#SPJ4

You might be interested in

2- A student ran 135 meters in 15 seconds. What was the student's velocity?

RoseWind [281]

Answer:

9 Brainly hahaha ............huh

4 0

3 years ago

A potter’s wheel moves from rest to an angular speed of 0.50 rev/s in 28.9 s. Assuming constant angular acceleration, what is it

Fudgin [204]

Answer:

0.108 rad/s².

Explanation:

Given that

Initial angular velocity ,ωi = 0 rad/s

Final angular velocity ωf= 0.5 rev/s

We know that

1 rev/s = 6.28 rad/s

ωf= 3.14 rad/s

t= 28.9 s

We know that (if acceleration is constant)

ωf=ωi + α t

α=Angular acceleration

3.14 = 0 + α x 28.9

$\alpha=\dfrac{3.14}{28.9}\ rad/s^2\\\alpha=0.108\ rad/s^2$

Therefore the acceleration will be 0.108 rad/s².

Therefore the answer will be 0.108 rad/s².

5 0

3 years ago

You have a source of energy containing 21 gj of energy at 600k how much this energy can be converted to work when rejecting heat

sweet [91]

Answer:

Available energy = 35 x 10⁶ J

Explanation:

Given:

Amount of energy (Q) = 21 gj = 21 x 10⁹ J

Temperature T1 = 600 k

Temperature T0 = 27 + 273 = 300k

Find:

Available energy

Computation:

Available energy = Q[1/T0 - 1/T1]

Available energy = 21 x 10⁹ J[1/300 - 1/600]

Available energy = 35 x 10⁶ J

4 0

3 years ago

Neutral substances have a pH of _____.

Alinara [238K]

Explanation:

Substances become neutral at the number 7 because at this point, the acid and Base are equal and becomes neutralized

4 0

3 years ago

0.180-kg stone rets on a frictionless, horizontal surface. A bullet of mass 7.50 g, traveling horizontally at 390 m/s, strikes t

Musya8 [376]

Answer:

The magnitud of the velocity is

$8.46m/s$

and the direccion:

$-28.3$ degrees from the horizontal.

Explanation:

Fist we define our variables:

$m_{s}=0.18kg\\m_{b}=7.5g = 0.0075kg\\v_{1b}= 390m/s -i\\v_{1s}=0m/s\\v_{2b}=210m/s -j\\v_{2s}=?$

The letters i and j represent the direction of the movement, i in this case is the horizontal direction, and j is perpendicular to i.

velocities with sub-index 1 are the speeds before the crash, and with sub-index 2 are the velocities after the crash.

Using conservation of momentum:

$m_{s}v_{1s}+m_{b}v_{1b}=m_{s}v_{2s}+m_{b}v_{2b}\\v_{1s}=0, so\\m_{b}v_{1b}=m_{s}v_{2s}+m_{b}v_{2b}$

Clearing for the velocity of the stone after the crash:

$v_{2s}=\frac{m_{b}v_{1b}-m_{b}v_{2b}}{m_{s}}$

Substituting known values:

$v_{2s}=\frac{0.0075kg(((390m/s-i)-210m/s-j)}{0.18kg}\\v_{2s}=(16.25m/s-i) - (8.75m/s-j)$

The magnitud of the velocity is :

$|v_{2s}|=\sqrt{(16.25m/s-i)^2 + (8.75m/s-j)^2}\\|v_{2s}|=18.46m/s$

and the direction:

$tan^{-1}(-8.75/16.25)=-28.3$

this is -28.3 degrees from the +i direction or the horizontal direcction.

Note: i and j can also be seen as x and y axis.

3 0

4 years ago