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Ugo [173]
2 years ago
11

The water table is _____. found in the unsaturated zone always in the same location the top layer of the saturated zone made of

nonporous soil and rock
Physics
2 answers:
algol [13]2 years ago
4 0

The water table is the top layer of the saturated zone.

Ivanshal [37]2 years ago
3 0

Answer:  the top layer of the saturated zone

Explanation: Water table is also called as the groundwater table.The water table is the level below the surface of the ground where water is found.

The water table is the top layer of the saturated zone.The saturated zone is where the pores and fractures of the ground are saturated with water.

The water table keeps changing with the seasons and from year to year because it is affected by climatic variations and by the amount of precipitation.

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In 3 seconds a car moving in a straight line increases its speed from 22.4 m/s to 29.1 m/s while a truck increases its speed fro
ANTONII [103]
Is there a picture?
6 0
3 years ago
Un auto de 2,300 kg está estacionado sobre una rampa inclinada 26.0°. Obtenga la tensión en la cadena.
Natasha_Volkova [10]

Answer:

T = mgsinθ = 2300(9.8)sin26.0 = 9880.88 ≈ 9900 N

Explanation:

8 0
2 years ago
After a check up, a person now has a far point of 100 cm, but with good near point vision. He needs to wear a new pair of correc
lakkis [162]

Answer:

so his far point according to this pair of glass is 200 cm

Explanation:

power of old pair of corrective glasses is given as

P = -0.5 dioptre

now we have

f = \frac{1}{P}

f = -2 m

f = -200 cm

now we know that for normal vision the maximum distance of vision is for infinite distance

so by lens formula we have

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

\frac{1}{d_i} + 0 = \frac{1}{200}

d_i = 200 cm

so his far point according to this pair of glass is 200 cm

7 0
3 years ago
car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of a. The car
Luba_88 [7]

Answer:

0.572

Explanation:

First examine the force of friction at the slipping point where Ff = µsFN = µsmg.

the mass of the car is unknown,

The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions.

First the tangential direction

∑Ft =Fft =mat

And then in the centerward direction ∑Fc =Ffc =mac =mv²t/r

Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2

So going backwards and plugging in Ffc =m2atπr/ 2r =πmat

Ff = √(F2ft +F2fc)= matp √(1+π²)

µs = Ff /mg = at /g √(1+π²)=

1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572

7 0
3 years ago
A 57.0 kg cheerleader uses an oil-filled hydraulic lift to hold four 120 kg football players at a height of 1.10 m. If her pisto
lapo4ka [179]

Answer:

The diameter of the piston of the players equals 55.136 cm.

Explanation:

from the principle of transmission of pressure in a hydraulic lift  we have

\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{1}}

Since the force in the question is the weight of the individuals thus upon putting the values in the above equation we get

\frac{57.0\times 9.81}{\frac{\pi \times (19.0)^{2}}{4}}=\frac{4\times 120\times 9.81}{\frac{\pi \times D_{2}^{2}}{4}}

Solving for D_{2} we get

D_{2}^{2}=\frac{4\times 120}{57}\times 19^{2}\\\\\therefore D_{2}=\sqrt{\frac{4\times 120}{57}}\times 19\\\\D_{2}=55.136cm

5 0
3 years ago
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