What happens to the mass and volume if density increases

A well is pumped at Q = 300 m3 /hr in a confined aquifer. The aquifer transmissivity is 25 m2 /hr and the storage coefficient is

Effectus [21]

Answer:

(a). The draw-down at a distance 200 m from the well after pumping for 50 hr is 5.383 m.

(b). The draw-down at a distance 200 m from the well after pumping for 50 hr is 6.707 m.

Explanation:

Given that,

Energy $Q=300\ m^3/hr$

Transmissivity $T = 25\ m^2/hr$

Storage coefficient $S=2.5\times10^{-4}$

Distance r= 200 m

We need to calculate the draw-down at a distance 200 m from the well after pumping for 50 hr

Using formula of draw-down

$s= \dfrac{Q}{4\pi T}(-0.5772-ln\dfrac{r^2S}{4tT})$

Put the value into the formula

$s=\dfrac{300}{4\pi\times25}(-0.5772-ln\dfrac{(200)^2\times2.5\times10^{-4}}{4\times25\times50})$

$s=5.383\ m$

We need to calculate the draw-down at a distance 200 m from the well after pumping for 200 hr

Using formula of draw-down

$s= \dfrac{Q}{4\pi T}(-0.5772-ln\dfrac{r^2S}{4tT})$

Put the value into the formula

$s=\dfrac{300}{4\pi\times25}(-0.5772-ln\dfrac{(200)^2\times2.5\times10^{-4}}{4\times25\times200})$

$s=6.707\ m$

Hence, (a). The draw-down at a distance 200 m from the well after pumping for 50 hr is 5.383 m.

(b). The draw-down at a distance 200 m from the well after pumping for 200 hr is 6.707 m.

3 0

3 years ago

Which change of state has the wrong energy change listed? condensation deposition melting freezing

ipn [44]

Where are the energy changes?

6 0

3 years ago

Read 2 more answers

EXPERIMENT:ELECRIC MOTOR.

g100num [7]

Answer:

1. Current flows through the battery, aluminum foil, and paper clips, and into the wire coil, creating an electromagnet. One face of the coil becomes a north pole; the other a south pole. The permanent magnet attracts its opposite pole on the coil and repels its like pole, causing the coil to spin.

- Quoted from Google

2. Electrical engineers say that, in an electrical circuit, electricity flows one direction: out of the positive terminal of a battery and back into the negative terminal. Electronic technicians say that electricity flows the other direction: out of the negative terminal of a battery and back into the positive terminal.

- Quoted from Google

7 0

3 years ago

An airplane is flying with a velocity of 100 m/s at an angle of 25° above the horizontal. When the plane is 114 m directly above

alex41 [277]

Answer:

The suitcase will land 976.447m from the dog.

Explanation:

The velocity in its component in the X and Y axis is decomposed:

Vx= 100m/s × cos(25°)= 90.63m/s

Vy= 100m/s × sen(25°)= 42.26m/s

Time it takes for the suitcase to reach maximum height, the final speed on the axis and at the point of maximum height is zero whereby:

VhmaxY= Voy- 9.81(m/s^2) × t ⇒ t= (42.26 m/s) / (9.81(m/s^2)) = 4.308s

The space traveled on the axis and from the moment the suitcase is thrown until it reaches its maximum height will be:

Dyhmax= Voy × t - (1/2) × 9.81(m/s^2) × (t^2) =

= 42.26m/s × 4.308s - 4.9 (m/s^2) × (4.308s)^2 =

=182.056m - 90.938m= 91.118m

The time from the maximum height to touching the ground is:

Dtotal y= 114m + 91.118m = (1/2) × 9.81(m/s^2) × (t^2) =

= 205.118m = 4.9 (m/s^2) × (t^2) ⇒ t= (41.818 s^2) ^ (1/2)= 6.466s

The total time of the bag in its rise and fall will be:

t= 4.308s + 6.466s = 10.774s

With this time and the initial velocity at x which is constant I can obtain the distance traveled by the suitcase on the x-axis:

Dx= 90.63 (m/s) × 10.774s = 976.447m

8 0

3 years ago

Two parallel-plate capacitors, 6.0 mF each, are connected in parallel to a 10 V battery. One of the capacitors is then squeezed

Lady_Fox [76]

The additional charge transferred to the capacitors by the battery is 60 μC.

The increase in the total charge stored on the capacitors is 60 μC.

<u>Explanation</u>:

C = εоA / d

If the separation is halved, then the capacitance will be doubled and according to the equation q = CV, the charge will be doubled too.

Initial charge of the capacitor is:

q = CV

= (6e - 6) $\times$ (10)

= 60 μC

Final charge of the capacitor:

q = (2C)V

= (2 $\times$ 6e - 6) $\times$ (10)

= 120 μC

additional charge transmitted is:

q' = 120 - 60

= 60 μC

initial total charge:

$q_{i}$ = (C1 + C2) V

= (6 + 6) $\times$ (10)

= 120 μF

final total charge:

$q_{f}$ = (C1 + C2) V

= (2 $\times$ 6 + 6) $\times$ (10)

= 180 μF

Increase in the charge:

q' = 180 - 120

= 60 μC

6 0

3 years ago