Answer:
<em>likely to decrease downstream in arid regions and increase downstream in temperate regions</em>
<em></em>
Explanation:
Arid regions are is a region with a severe lack of water, usually to the extent that affect the organisms living in the region. Arid regions are characterized by a very low depth of rainfall per year. Temperate region on the other hand experience more distinct seasonal change and wider temperature change. Temperate regions get a fairly large amount of rainfall per year.
In arid regions, the soil is very dry, and the rate of infiltration and percolation is high relative to the amount of rainfall available. The effect is that more water is infiltrated into the soil as you move downstream, leading to a decrease in the discharge of a stream as you move downstream. Most temperate region have soils that are usually saturated in the peak of the rainfall season, leading to a greater stream discharge as you move downstream.
Answer:
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Explanation:
Answer:
true
Explanation:
I'm assuming this is a true or false question. if u get the same results after each repeated trial that is the only time a hypothesis can be used to support evidence
Momentum would be the same before and after the collision
Before the collision:
Momentum of the single cart: 1 * 0.50 = 0.50
After the collision
velocity = 0.25m / s
1 * 0.25 + 1 * 0.25 =
0.25 * (1 + 1) =
0.25 * 2 =
0.50
Now new momentum will be 0.5
answer
the same before and after the collision
Answer:
Part(a): the capacitance is 0.013 nF.
Part(b): the radius of the inner sphere is 3.1 cm.
Part(c): the electric field just outside the surface of inner sphere is 
.
Explanation:
We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and '
' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as
Part(a):
Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.
So the capacitance (C) of the shell is
Part(b):
Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,
Part(c):
If we apply Gauss' law of electrostatics, then
