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Mariulka [41]
2 years ago
6

Tại sao bán cầu Nam lại mưa nhiều hơn bán cầu Bắc?

Geography
1 answer:
umka21 [38]2 years ago
7 0

Answe

rta/¿=Pqué llueve más el hemisferio sur que el hemisferio norte?

Razones por las que llueve más en el hemisferio norte

Circulación de los océanos será el principal motor de la asimetría en el modo de lluvia tropical lluvias en el norte que en el sur de la línea ecuatorial, sin embargo, el hemisferio sur expuesto más con la radiación solar, y por lo tanto a una mayor tasa de aumento de la humedad.

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The Resource Conservation and Recovery Act of 1976 resulted in _____.
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The correct answer is<span> A. a decrease in the dumping of hazardous waste

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When the sun produces energy, it consumes hydrogen and produces
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. If the fastest passenger aircraft can fly 0.45 km/s (1000 mph), how long would it take to reach the sun? The galactic center?
nadezda [96]

Answer:

The passenger aircraft would take 10.542 years to reach the Sun from the Earth.

The passenger aircraft would take 1.733\times 10^{10} years to reach the gallactic center.

Explanation:

The distance to the sun from the Earth is approximately equal to 1.496\times 10^{8}\,km, if the passenger travels at constant speed, then the time needed to reach the sun is calculated by the following kinematic formula:

\Delta t = \frac{s}{v} (1)

Where:

s - Travelled distance, measured in kilometers.

v - Speed of the passenger aircraft, measured in kilometers per second.

\Delta t - Travelling time, measured in seconds.

If we know that s = 1.496\times 10^{8}\,km and v = 0.45\,\frac{km}{s}, then the travelling time is:

\Delta t = \frac{1.496\times 10^{8}\,km}{0.45\,\frac{km}{h} }

\Delta t = 3.324\times 10^{8}\,s

\Delta t = 3847.736\,days

\Delta t = 10.542\,years

The passenger aircraft would take 10.542 years to reach the Sun from the Earth.

The distance between the Earth and the galactic center is approximately equal to 2.460\times 10^{17}\,km. If the passenger travels at constant speed and if we know that s = 2.460\times 10^{17}\,km and v = 0.45\,\frac{km}{s} , then the travelling time is:

\Delta t = \frac{2.460\times 10^{17}\,km}{0.45\,\frac{km}{s} }

\Delta t = 5.467\times 10^{17}\,s

\Delta t = 6.327\times 10^{12}\,days

\Delta t = 1.733\times 10^{10}\,years

The passenger aircraft would take 1.733\times 10^{10} years to reach the gallactic center.

3 0
3 years ago
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