The point is the water table closest to the surface in valleys.
In areas of topographic relief, the water table generally follows the surface but tends to approach it in valleys and intersect the surface with streams and lakes. Closest to the surface is the vented zone where the interstices between the soil are filled with both air and water. Below this layer is a saturation zone where the gaps are filled with water.
The water table is the subsurface boundary between the soil surface and the area where groundwater saturates the space between sediment and rock fissures. At this boundary, water pressure equals atmospheric pressure. The top surface of the groundwater is the groundwater table. Beneath this surface, all pore spaces and cracks in sediments and rocks are completely filled and saturated with water. Groundwater occurs in these saturated layers known as saturation zones or water vapor zones.
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Answer:
The distance traveled by the woman is 34.1m
Explanation:
Given
The initial height of the cliff
yo = 45m final, positition y = 0m bottom of the cliff
y = yo + ut -1/2gt²
u = 20.0m/s initial speed
g = 9.80m/s²
0 = 45.0 + 20×t –1/2×9.8×t²
0 = 45 +20t –4.9t²
Solving quadratically or by using a calculator,
t = 5.69s and –1.61s byt time cannot be negative so t = 5.69s
So this is the total time it takes for the ball to reach the ground from the height it was thrown.
The distance traveled by the woman is
s = vt
Given the speed of the woman v = 6.00m/s
Therefore
s = 6.00×5.69 = 34.14m
Approximately 34.1m to 3 significant figures.
As we know that centripetal force =mv^2/r
given data is
m = mass
v = speed
r = radius
putting values we get
= 85 x 15^2 / 20
= 956.25 N
option d is correct
Refer to the diagram shown below.
The initial KE (kinetic energy) of the system is
KE₁ = (1/2)mu²
After an inelastic collision, the two masses stick together.
Conservation of momentum requires that
m*u = 2m*v
Therefore
v = u/2
The final KE is
KE₂ = (1/2)(2m)v²
= m(u/2)²
= (1/4)mu²
= (1/2) KE₁
The loss in KE is
KE₁ - KE₂ = (1/2) KE₁.
Conservation of energy requires that the loss in KE be accounted for as thermal energy.
Answer: 1/2
Answer:
A) 185.6 J
B) 9.396 x 10^14 J
C) 4x10^7 m/s
D) 20 m
E) 9.09x10^-8 sec
F) 9.09x10^-8 sec
Explanation:
Detailed explanation and calculation is shown in the image below
