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3 years ago

Population Pyramids

1 answer:

7 0

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The climate of a city near the equator is warmer than the climate of a

ss7ja [257]

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Places near the equator face the sun more often and therfore are warmer.

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3 years ago

Read the excerpt from We Shall Not Be Moved. Publicity was another vital need — photos and eyewitness accounts of workers on the

Gwar [14]

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3 years ago

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Political and economic systems <br>features of goverment

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Political system. A political system is a system of politics and government. It is usually compared to the legal system, economic system, cultural system, and other social systems.

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4 years ago

What motivated Europeans and Africans to turn to selling people as property during the Age of Exploration?

shusha [124]

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Explanation: The purpose of the exploration was to expand European geographic knowledge, to find the west of Africa for example, slaves could marry, own property and even own slaves

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4 years ago

. 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.

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3 years ago