What is the objective of a geopark

1 answer:

3 0

~The <span>Drive landscape-scale conservation of geo-diversity, biodiversity and cultural heritage.
~Enthuse local communities and their businesses to work together sustainably and feel pride in the Geopark.
~Promote sustainable economic development, especially sustainable tourism.
~Develop high quality information and interpretation that supports an enjoyable visitor experience and connects people to the Geopark.
~Act as a focus for environmental/earth sciences research.
~Undertake high quality environmental education for all ages.
~Seek to include all sectors of society in developing its work.
~Use its heritage to encourage visitors, businesses and communities to address climate change.
<span>~Monitor impacts of the Geopark in order to inform future actions.</span></span>

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What type of environment is shown in the image?

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Read 2 more answers

. 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: