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irina [24]
2 years ago
9

Part B Next you'll investigate how precipitation varies throughout Earth. Visit the My NASA Data e website and select Data Set i

n the top left corner. Within the Hydrosphere category, select All Data. Within All data category select Precipitation, and then select Monthly Precipitation. After the new map appears, select Animate near the top of the page, and then select OK to see the animation. Set the time range on the animation for the most recent five years, and then click Submit. You can control the speed of the animation with the controls beneath the map. Question 1 Describe patterns you see in the precipitation on Earth. How do these patterns change during the course of a year and from year to year? Use lines of latitude and the amount of precipitation in millimeters/day (mm/day) in your description.​
Physics
1 answer:
Lera25 [3.4K]2 years ago
6 0

The patterns you see in the precipitation on Earth are that it is often distributed evenly or unevenly via:

  • Geographical method
  • Temporal method
  • Seasonal method

<h3>What is the pattern of precipitation on earth?</h3>

The yearly precipitation over the Earth is found to be approximately about 100 cm (that is 39 inches).

Note that is one that is often distributed unevenly. The areas of highest rainfall is known to be in the equatorial zone and the monsoon area of Southeast Asia.

Learn more about precipitation from

brainly.com/question/1783904

#SPJ1

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1. Physical size of Russia compared to other countries, despite a lack of visible borders from space.

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7 0
3 years ago
An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 10
s2008m [1.1K]

Answer:

<em>1.228 x </em>10^{-6}<em> mm </em>

<em></em>

Explanation:

diameter of aluminium bar D = 40 mm  

diameter of hole d = 30 mm

compressive Load F = 180 kN = 180 x 10^{3} N

modulus of elasticity E = 85 GN/m^2  = 85 x 10^{9} Pa

length of bar L = 600 mm

length of hole = 100 mm

true length of bar = 600 - 100 = 500 mm

area of the bar A = \frac{\pi D^{2} }{4} =  \frac{3.142* 40^{2} }{4} = 1256.8 mm^2

area of hole a = \frac{\pi(D^{2} - d^{2}) }{4} = \frac{3.142*(40^{2} - 30^{2})}{4} = 549.85 mm^2

Total contraction of the bar = \frac{F*L}{AE} + \frac{Fl}{aE}

total contraction = \frac{F}{E} * (\frac{L}{A} +\frac{l}{a})

==> \frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85}) = <em>1.228 x </em>10^{-6}<em> mm </em>

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