The bulk of the world's deserts are located at 30 degrees north latitude and 30 degrees south latitude, when the warm equatorial air begins to descend. The heavy, warm, descending air vaporises large amounts of water from the ground's surface. As a result, the environment is rather dry.
<h3>Why are the majority of the desert regions on Earth located between 20 and 30 degrees latitude?</h3>
The zones of falling air are those between 20 and 30 latitudes on the western borders of continents (high pressure and dry weather). As a result, the moisture continues to decrease as the air is compressed and warmed as it falls.
Where the scorching equatorial air starts to descend, the majority of the world's deserts are found between 30 degrees north and 30 degrees south latitude. Large volumes of water are vaporised off the surface of the ground by the thick, warming, falling air. As a result, the climate is extremely dry.
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Answer:
A) -2N
B) Left
C) -0.5
Explanation:
A) -12 + 10
B) More force is acted on in that direction
C) Net force/Mass (-2/4)
C. meter per second
Velocity and speed share the same SI unit.
The partial pressure of the O2 is 36.3 kiloPascal when the air pressure in the mask is 110 kiloPascal based on the isotherm relation. This problem can be solved by using the isotherm relation equation which stated as Vx/Vtot = px/ptot, where V represents volume, p represents the pressure, x represents the partial gas, and tot represents the total gas<span>. Calculation: 33/100 = px/110 --> px = 36.3</span>
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vo = 25 m/sec
<span>vf = 0 m/sec </span>
<span>Fμ = 7100 N (Force due to friction) </span>
<span>Fg = 14700 N </span>
<span>With the force due to gravity, you can find the mass of the car: </span>
<span>F = ma </span>
<span>14700 N = m (9.8 m/sec²) </span>
<span>m = 1500 kg </span>
<span>Now, we can use the equation again to find the deacceleration due to friction: </span>
<span>F = ma </span>
<span>7100 N = (1500 kg) a </span>
<span>a = 4.73333333333 m/sec² </span>
<span>And now, we can use a velocity formula to find the distance traveled: </span>
<span>vf² = vo² + 2a∆d </span>
<span>0 = (25 m/sec)² + 2 (-4.73333333333 m/sec²) ∆d </span>
<span>0 = 625 m²/sec² + (-9.466666666667 m/sec²) ∆d </span>
<span>-625 m²/sec² = (-9.466666666667 m/sec²) ∆d </span>
<span>∆d = 66.0211267605634 m </span>
<span>∆d = 66.02 m</span>
