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Mila [183]
3 years ago
6

What conditions are necessary for an artesian well

Physics
1 answer:
madreJ [45]3 years ago
7 0
The geologic<span> conditions necessary for an artesian well are an inclined aquifer sandwiched between impervious </span>rock layers<span> above and below that trap water in it. Water enters the exposed edge of the aquifer at a high </span>elevation<span> and percolates downward through interconnected </span>pore spaces<span>. </span>
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the displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s
12345 [234]

The average velocity or displacement of a particle for the first time interval is <u>Δs / Δt = 6 cm/s.</u>

Solution:

As we know that displacement is calculated in centimeters and the unit of time is second.

The average velocity for the first interval [1,2] is given

Δs / Δt = s (t2) - s (t) / t2 - t1

Δs / Δt = 2sin2  π  + 3cos 2 π -  ( 2sin π + 3cos π ) / 2 - 1

Δs / Δt = 2(0) + 3(1) - 2(0) - 3 (-1) / 1

Δs / Δt = 6 cm/s

Thus the average velocity or displacement of a particle for the first time interval is Δs / Δt = 6 cm/s

If you need to learn more about displacement click here:

brainly.com/question/28370322

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The complete question is:

The displacement of a particle moving back and forth along a line is given by the following equation s(t) = 2sin π t + 3cos π t. Estimate the instantaneous velocity of the particle when t = 1

4 0
1 year ago
Volcanoes change the earth by
SOVA2 [1]
I believe the answer is b) slowly heating the surface
8 0
2 years ago
Read 2 more answers
Two streams merge to form a river. One stream has a width of 8.4 m, depth of 3.5 m, and current speed of 2.2 m/s. The other stre
hoa [83]

Answer:

Explanation:

We shall solve this problem on the basis of pinciple that water is incompressible so volume of flow will be equal at every point .

rate of volume flow of one stream

= cross sectional area x velocity

= 8.4 x 3.5 x 2.2 = 64.68 m³ /s

rate of volume flow of other stream

= 6.6 x 3.6 x 2.7

= 64.15 m³ /s

rate of volume flow of rive , if d be its depth

= 11.2 x d x 2.8

= 31.36 d

volume flow of river = Total of volume flow rate of two streams

31.36 d  = 64.15 + 64.68

31.36 d  = 128.83

d = 4.10 m /s .

6 0
2 years ago
Part B
Licemer1 [7]

The number of kilowatts used by an individual to operate his appliances is determined as 12.1 kWh.

<h3>Average daily power consumption</h3>

The average daily power consumption is the amount of electric energy consumed by an individual on a daily rate.

The average daily power consumption of individuals in USA is 12,100 W-hr.

<h3>Converting watts to kilowatts</h3>

E = 12,100 Whr/1000

E = 12.1 kWh

Thus, the number of kilowatts used by an individual to operate his appliances is determined as 12.1 kWh.

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5 0
2 years ago
1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x
allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}}  }

With (\epsilon_{0}) been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} }    }

Explanation:

(A) Considering a uniform linear density λ_{0} on the ring, then:

dQ=\lambda dl (1)⇒Q=\lambda_{0} 2\pi R(2)⇒\lambda_{0}=\frac{Q}{2\pi R}(3)

Applying the technique of charge integration for finite charges:

V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r'  }} \, dQ(4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

r'=\sqrt{R^{2} +Z^{2}}(5)

Using the expressions (1),(4) and (5) you obtain:

V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}}  }} \, d\phi

Integrating results:

V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}}  }   (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

|E| =-\nabla V (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

|E| =-\frac{dV(z)}{dz} (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} }    } (S_b)

4 0
3 years ago
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