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cupoosta [38]
2 years ago
8

is a climate cycle where the temperature of the Ocean changes because of movements of air and ocean currents. During this period

, the trade winds in the south , causing the water to move from the region to the region.
Physics
1 answer:
k0ka [10]2 years ago
3 0
<h2>Answer: <u>El Niño</u> phenomenon </h2>

<em>El Niño</em> is a natural climatic phenomenon that occurs cyclically on our planet Earth (albeit irregularly) in periods of between 3 and 7 years. It occurs when the temperature of the Pacific Ocean in its eastern part, which is towards South America, becomes warmer than usual.

To understand it better, it is necessary to know the following:

Normally in the referred area of the Pacific, trade winds blow from east to west, which move the warm water from the area to the west, allowing deeper and colder water to emerge. But, during El Niño the trade winds weaken or even flow in reverse, from west to east. It is then, when the warm water accumulates in front of the pacific coasts of South America provoking a change in the habitual patterns of precipitation and temperature.

As a consequence, hot water evaporates and condenses, causing an increase in rainfall. Due to this change in the trade winds, the climate pattern is altered, and these rains affect more the eastern part of the Pacific, while at the other end of the ocean (in Australia and Southeast Asia) the climate becomes colder and dry.

It should be noted that this is a global phenomenon that affects and extends directly or indirectly in most regions of the planet. However, its greatest impact occurs in the American countries of the Pacific coast; and in Southeast Asia and Australia.

This phenomenom begins during the final months of the year, from October and generally may last until January of the following year, although its effects are often maintained even until March.

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An electron passes location &lt; 0.02, 0.04, -0.06 &gt; m and 5 us later is detected at location &lt; 0.02, 1.62,-0.79 &gt; m (1
hram777 [196]

Answer:

(a)

Average velocity = < 0, 316000, 146000> m/s

(b) < 0, 2.844, 1.314 > m

Explanation:

r1 = < 0.02, 0.04, - 0.06 > m

r2 = < 0.02, 1.62, - 0.79 > m

time, t = 5 micro second = 5 x 10^-6 s

(a) Average velocity is defined as the ratio of total displacement to the total time taken.

Displacement = r = r2 - r1

r = < 0.02 - 0.02, 1.62 - 0.04, - 0.79 + 0.06 > m

r = < 0, 1.58, - 0.73 > m

So. Average velocity =  \frac{< 0, 1.58, - 0.73 >}{5 \times 10^{-6}}

Average velocity = < 0, 316000, 146000> m/s

Average velocity = < 0, 316000, 146000> m/s

(b)

Distance = velocity x time

Here time, t = 9 micro second

d =  < 0, 316000, 146000>  x 9 x 10^-6 m

d = < 0, 2.844, 1.314 > m

5 0
3 years ago
How far should the second antenna be moved in order to receive a minimum signal from a station that broadcasts at 99.4 MHz?
raketka [301]

Answer:

So the distance of the antenna from the station will be 3.018 m

Explanation:

We have given the frequency of the broadcast f=99.4MHz=99.4\times 10^4Hz

The speed of light c=3\times 10^8m/sec

The distance of the antenna to receive a minimum signal from the station is given by d=\frac{v}{f}=\frac{3\times 10^{8}}{99.4\times 10^6}=3.018m

So the distance of the antenna from the station will be 3.018 m

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3 years ago
15 points. give me the method.
AveGali [126]

Answer:

\boxed{{160 \:  m(s)}^{ - 1} }

Explanation:

if \:the \:  frequencies \: are \to \\   f_{1} =  640Hz  \\ and \\f_{2}   = 480Hz \:  \\ but \:  \boxed{v = f \gamma }:   f =  \frac{v}{ \gamma } \\ if \:  \gamma_{1}  -  \gamma _{2}  = 1 =  \gamma  \\ f_{1}  - f_{2}  = 640 - 480 = \boxed{ 160Hz} = f \\ v = f \gamma = 160 \times 1 =  \boxed{{160 \:  m(s)}^{ - 1} }

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