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3 years ago

How long do molecules of groundwater stay in the ground?

Physics

2 answers:

Hoochie [10]3 years ago

6 0

Answer:

It varies from days to years

Explanation:

Water cycle includes series of steps and processes which enables water to be recycled. During dry or hot season , the sun heats up water bodies and water evaporates into the atmosphere.

During the cold or rainy season the evaporated water condensed and fall as rain back into the water bodies and environment generally.

The period or time frame in which ground water from rain stays in the soil is based on some factors such as the nature of the soil. Some soils such as clayey soil have a larger water retentive capacity than the others.

Brilliant_brown [7]3 years ago

4 0

CORRECT ANSWER:

d. Anywhere from days to thousands of years.

STEP-BY-STEP EXPLANATION:

The whole question from book is

How long do molecules of groundwater stay in the

ground?

a. Days

b. Weeks

c. Months

d. Anywhere from days to thousands of years

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Two objects carry initial charges that are q1 and q2, respectively, where |q2| > |q1|. They are located 0.160 m apart and beh

mart [117]

Answer:

$\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.$

Explanation:

According to the Coulomb's law, the magnitude of the electrostatic force between two static point charges $\rm q_1$ and $\rm q_1$ , separated by a distance $\rm r$ , is given by

$\rm F = \dfrac{kq_1q_2}{r^2}.$

where k is the Coulomb's constant.

Initially,

$\rm r = 0.160\ m\\F_i = -1.30\ N.\\\\and \ \ |q_2|>|q_1|.$

The negative sign is taken with force F because the force is attractive.

Therefore, the initial electrostatic force between the charges is given by

$\rm F_i = \dfrac{kq_1q_2}{r^2}.\\-1.30=\dfrac{kq_1q_2}{0.160^2}\\\rm\Rightarrow q_2 = \dfrac{-1.30\times 0.160^2}{q_1k}\ \ \ ..............\ (1).$

Now, the objects are then brought into contact, so the net charge is shared equally, and then they are returned to their initial positions.

The force is now repulsive, therefore, $\rm F_f = +1.30\ N.$

The new charges on the two objects are

$\rm q_1'=q_2' = \dfrac{q_1+q_2}{2}.$

The new force is given by

$\rm F_f = \dfrac{kq_1'q_2'}{r^2}\\+1.30=\dfrac{k\left (\dfrac{q_1+q_2}{2}\right )\left (\dfrac{q_1+q_2}{2}\right )}{0.160^2}\\\Rightarrow \left (\dfrac{q_1+q_2}{2}\right )^2=\dfrac{+1.30\times 0.160^2}{k}\\(q_1+q_2)^2=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+q_2^2+2q_1q_2=\dfrac{4\times 1.30\times 0.160^2}{k}\\\\$

Using (1),

$\rm q_1^2+\left ( \dfrac{-1.30\times 0.160^2}{q_1k}\right )^2+2\left (\dfrac{-1.30\times 0.160^2}{k} \right )=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+\dfrac 1{q_1^2}\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0$

$\rm q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )=0\\q_1^4-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2=0$

$\rm q_1^4-q_1^2\left (2.22\times 10^{-11} \right )+\left ( 1.37\times 10^{-23}\right ) =0\\\Rightarrow q_1^2 = \dfrac{-(-2.22\times 10^{-11})\pm \sqrt{(-2.22\times 10^{-11})^2-4\cdot (1)\cdot (1.37\times 10^{-23})}}{2}\\=1.11\times 10^{-11}\pm 1.046\times 10^{-11}.\\=6.4\times 10^{-13}\ \ \ or\ \ \ 2.156\times 10^{-11}\\\Rightarrow q_1 = \pm 8.00\times 10^{-7}\ C\ \ \ or\ \ \ \pm 4.64\times 10^{-6}\ C.$

Using (1),

When $\rm q_1 = \pm 8.00\times 10^{-7}\ C$ ,

$\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 8.00\times 10^{-7}\times 9\times 10^9}=\mp4.6\times 10^{-6}\ C.$

When $\rm q_1=\pm 4.6\times 10^{-6}\ C$ ,

$\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 4.64\times 10^{-6}\times 9\times 10^9}=\mp7.97\times 10^{-7}\ C\approx 8.0\times 10^{-7}\ C.$

Since, $\rm |q_2|>|q_1|$

Therefore, $\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.$

7 0

2 years ago

What is the area of the circle with a radius of 12.77m

Arada [10]

A=pi r^2
A=( 3.14159.......)(12.77)^2
A= 512.30m^2

8 0

3 years ago

A 9Ω resistance is cut into three equal parts and connected in parallel. Find the equivalent resistance of the combination.

sammy [17]

Answer:

the ans i 1 ohm

Explanation:

if i cut a resistor of 9 ohm into 3 equal parts then each resistor will have a resistance of 3 ohm and if they are in parallel combination the net resistance will be R

so 1/R=1/R1+1/R2+1/R3

1/R=1/3+1/3+1/3

R=1

7 0

2 years ago

A working fluid enters a steady flow system with a velocity of 30m/s and leaves at

Oxana [17]

Answer:

Pretty sure it’s A

Explanation:

Would ike brainliest

6 0

3 years ago

1 A standard transformer has a rating of 60 kVA. The primary and secondary voltages are rated respectively at 600 V and 120 V. (

Viefleur [7K]

Answer:

The rated current for the 600V side is I = 100 A and the rated current for the 120 V side is 500 A.

Explanation:

For standard transformers they step up or down the level of voltage while maintaning the power output (ideally). So the same power output should be seen on both sides, since the equation for apparent power is P = V*I we can solve for I on both sides and find the rated current. So we have:

600 V side:

P = V*I

I = P/V = (60*10^3)/600 = 100 A

120 V side:

P = V*I = (60*10^3)/120 = 500 A

6 0

3 years ago

Read 2 more answers