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1 year ago

At which point is the water table closest to the surface? Explain.

Physics

1 answer:

dangina [55]1 year ago

7 0

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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The human nerve cells have a net negative charge and the material in the interior of the cell is a good conductor. if the cell h

mash [69]

The magnitude and direction (inward or outward) of the net flux through the cell boundary is - 0.887 wb.m².

Flux describes any effect that appears to pass or travel through a surface or substance.

The magnitude and direction (inward or outward) of the net flux through the cell boundary is calculated as follows;

Ф = Q/ε

where;

Q is net charge
ε is permittivity of free space

Φ = (-7.85 x 10⁻¹²)/(8.85 x 10⁻¹²)

Φ = - 0.887 wb.m²

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6 0

2 years ago

A race car travels 40 m/s around a banked (45° with the horizontal) circular (radius = 0.20 km) track. What is the magnitude of

OLEGan [10]

Answer:

c) $F_{net} = 0.640 kN$

Explanation:

As we know that resultant force is the net force that is acting on the system

As per Newton's II law we know that net force is product of mass and acceleration

so we will have

$F_{net} = ma$

here we know

m = 80 kg

for circular motion acceleration is given as

$a_c = \frac{v^2}{R}$

$a_c = \frac{40^2}{200} = 8 m/s^2$

now we have

$F_{net} = 80 \times 8$

$F_{net} = 640 N$

$F_{net} = 0.640 kN$

7 0

3 years ago

A car changes speed from 25 m/s to 10 m/s in 240 seconds. Describe its acceleration.

Dmitry_Shevchenko [17]

Explanation:

Acceleration is change in velocity over change in time:

a = Δv / Δt

a = (10 m/s - 25 m/s) / (240 s - 0 s)

a = -0.0625 m/s²

So the car decelerates at 0.0625 m/s².

7 0

4 years ago

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$