What type of cloud is highest in the sky?

Rogue waves are created along the "Wild Coast" off the southeast coast of ________, where the Agulhas Current flows directly aga

Vedmedyk [2.9K]

Answer:

Africa

Explanation:

A rogue wave refers to the wave that is twice the height of a significant wave occurring in a particular area. The significant wave height is generally referred to as the mean of the largest one-third of waves existing at a particular time period. In simple words, a rogue wave is much larger than any other waves that occur at the proximity of the same time.

This rough wave describes the interaction between the ocean and sea current and swelling of waves. It takes place when the large swells in the ocean, also known as the Antarctic storms, strikes with the rapidly traveling Agulhas current, and the curved water current focuses on the energy of the waves.

Thus, these Rogue waves are often generated along the southeastern coastal regions of Africa, where there occurs the convergence of Antarctic storm waves and Agulhas Current.

4 0

3 years ago

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t

givi [52]

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

$f = f_0 (\frac{v_0}{v_0-v})$

Here,

$f_0$ = Frequency of Source

$v_s$ = Speed of sound

f = Frequency heard before slowing down

f' = Frequency heard after slowing down

v = Speed of the train before slowing down

So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,

The first equation is,

$f = f_0 (\frac{v_0}{v_0-v})$

$300 = f_0 (\frac{343}{343-v})$

$(300*343) - 300v = 343f_0$

Now the second expression will be,

$f' = f_0 (\frac{v_0}{v_0-v/2})$

$290 = (343)(\frac{v_0}{343-v/2})$

$290*343-145v = 343f_0$

Dividing the two expression we have,

$\frac{(300*343) - 300v}{290*343-145v} = 1$

Solving for v, we have,

$v = 22.12m/s$

Therefore the speed of the train before and after slowing down is 22.12m/s

6 0

3 years ago

Two froghoppers sitting on the ground aim at the same leaf, located 35 cm above the ground. Froghopper A jumps straight up while

evablogger [386]

Frog hopper B❤️recent emojis

3 0

3 years ago

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A car of mass 750 kg accelerates away from traffic lights. At the end of the first 100 m it has reached a speed

malfutka [58]

the work done on the car by the force of its engine is 78,000 J.

" The work done on the car by the force of friction is 24,000 J.

Increasing the car's kinetic energy at the end of the first 100 m is 54,000J

a. Completed work = force x distance. Engine output = 780 N, that is,

780 N x 100 m = 78,000 J.

b. Completed work = force x distance. Friction force = 240 N, that is,

240 N x 100 m = 24,000 J.

c. Kinetic energy = 1 \ 2 x m x v2

= 1 \ 2 x 750 kg x 12 squared = 375 x 144 = 54,000 J.

<h3>How powerful is the engine of a car? </h3>

Mainstream car and truck engines typically produce 100-400 pounds. -Torque feet. This torque is generated by the engine piston as it moves up and down on the engine crankshaft, causing the engine to rotate (or twist) continuously.

Learn more about work done here: brainly.com/question/25573309

#SPJ10

5 0

2 years ago

How can you determine the number of valence electrons in atom without drawing a Bohr model?

grigory [225]

Using your periodic table if you look at it 3-11 are tansition metals so the horizontal Group Number will help if the group number has to digits just remove the one so if it were to be 13, the valence would be 3, if it were 14 the valence would be ,4 if it were 15, the valence would be 5, if it were 16 the valence would be 6, if it were 17 the valence would be 7 if it were group 18 the valence would be 8 so if anymore help needed to explain hit me up

5 0

3 years ago

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