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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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RSB [31]

Answer:

The mass of the copper is 3.00kg

Explanation:

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2 years ago

A 15 kg runaway grocery cart runs into a spring with spring constant 230 N/m and compresses it by 56 cm .What was the speed of t

liubo4ka [24]

To solve this problem we will apply the concepts related to the conservation of kinetic energy and elastic potential energy. Thus we will have that the kinetic energy is

$KE = \frac{1}{2} mv^2$

And the potential energy is

$PE = \frac{1}{2} kx^2$

Here,

m = mass

v = Velocity

x = Displacement

k = Spring constant

There is equilibrium, then,

KE = PE

$\frac{1}{2} mv^2 = \frac{1}{2} kx^2$

Our values are given as,

$x=0.56m\\k=230N/m\\m=15kg$

Replacing we have that

$\frac{1}{2} (15)v^2 = \frac{1}{2} (230)(0.56)^2$

$v = 2.19m/s$

Therefore the speed of the cart is 2.19m/s

3 0

3 years ago

Someone pls help me with this!

Lina20 [59]

Answer:

Explanation:

<u>Coulomb's Law </u>

The electrostatic force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.

Written as a formula:

$\displaystyle F=k\frac{q_1q_2}{d^2}$

Where:

$k=9\cdot 10^9\ N.m^2/c^2$

q1, q2 = the objects' charge in coulomb

d= The distance between the objects in meters

Object 1 has a charge of

$q_1=-2.3\cdot 10^{-6}\ c$

Object 2 has a charge of

$q_2=-4.2\cdot 10^{-6}\ c$

They are separated by a distance of

d = 0.099 m

Calculate the force:

$\displaystyle F=9\cdot 10^9\frac{2.3\cdot 10^{-6}*4.2\cdot 10^{-6}}{0.099^2}$

F=8.87 N

5 0

2 years ago

If an object that weighs 20 N is lifted from the ground to a

Radda [10]

Answer:

false 20 n x 0.32 m = 6.4 J

3 0

3 years ago

Car A and Car B approach each other

vagabundo [1.1K]

The original frequency of horn of Car A is 1071 Hz.

Explanation:

Doppler effect describes the change in the frequency of sound waves with respect to the observer. As the sound waves emitted from a source need to travel the air medium to reach observer, it will undergo loss in energy. So there will be change in its frequency compared to original frequency. Depending upon the direction of travel of source and observer the shifting of frequency will vary.

$f'=\frac{v-v_{o} }{v-v_{s} } f$

Here vo is the observer velocity and vs is the velocity of the source. So Vo = 15 m/s as car B is the observer and Vs = 35 m/s as car A is the source. And f is the frequency of sound wave at source that is car A.

Similarly, the doppler shift in frequency is the frequency of sound heard by car B which is f' = 1140 Hz. And v is the speed of sound that is v = 343 m/s

1140 = $\frac{343-15}{343-35}*f= 1.0649 *f$

f = 1140/1.0649= 1071 Hz.

Thus, the original frequency of horn of Car A is 1071 Hz.

3 0

3 years ago