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3 years ago

ASAP I will give 34 points please give a good answer

Physics

2 answers:

liraira [26]3 years ago

6 0

Well, what's a well? Basically, to obtain water found in an aquifer, a well is a hole drilled into the earth. To pull water from the soil, a pipe and a pump are used and a screen filters out un Wells are drilled into the rock as deep as 1,000 feet to enter the water. ... Through a well pump, water flows through this casing. Above level, the well structure is closed off. From a pipe connected between the casing and a pressure tank (generally located in the basement of your house), the water then reaches your home.

yawa3891 [41]3 years ago

6 0

Answer:

This is an artesian well.

Explanation:

An artesian well is simply a well that doesn't require a pump to bring water to the surface. This occurs when there is enough positive pressure in the aquifer to bring the water to the surface. An artesian aquifer is confined between impermeable rocks or clay which causes this positive pressure.

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Using a radar gun, you emit radar waves at a frequency of 6.2 GHz that bounce off of a moving tennis ball and recombine with the

Keith_Richards [23]

Answer:

23.4 m/s

Explanation:

f = actual frequency of the wave = 6.2 x 10⁹ Hz

$f_{app}$ = frequency observed as the ball approach the radar

$f_{rec}$ = frequency observed as the ball recede away from the radar

V = speed of light

$v$ = speed of ball

B = beat frequency = 969 Hz

frequency observed as the ball approach the radar is given as

$f_{app}=\frac{f(V+v)}{V}$ eq-1

frequency observed as the ball recede the radar is given as

$f_{rec}=\frac{f(V-v)}{V}$ eq-2

Beat frequency is given as

$B = f_{app} - f_{rec}$

Using eq-2 and eq-1

$B = \frac{f(V+v)}{V}- \frac{f(V-v)}{V}$

inserting the values

$969 = \frac{(6.2\times 10^{9})((3\times 10^{8})+v)}{(3\times 10^{8})}- \frac{(6.2\times 10^{9})((3\times 10^{8})-v)}{(3\times 10^{8})}$

$v$ = 23.4 m/s

8 0

3 years ago

A uniform electric field of magnitude 375 n/c pointing in the positive x - direction acts on an electron, which is initially at

Finger [1]

(a) The force exerted by the electric field on the electron is given by the product between the electron charge q and the intensity of the electric field E:
$F=qE=(1.6 \cdot 10^{-19}C)(375 N/C)=6\cdot 10^{-17}N$
Under the action of this force, the electron moves by:
$\Delta x = 3.20 cm=0.032 m$
And the work done by the electric field on the electron is equal to the product between the magnitude of the force and the displacement of the electron. The sign has to be taken as positive, because the direction of the force is the same as the displacement of the electron, so:
$W=F \Delta x= (6\cdot 10^{-17}N)(0.032 m)=1.9 \cdot 10^{-18}J$

(b) The electron is initially at rest and it starts to move under the action of the electric field. This means that as it moves, it acquires kinetic energy and it loses potential energy. The change in potential energy is the opposite of the work done by the electric field:
$\Delta U = U_f - U_i = -1.9 \cdot 10^{-18} J$
Where Uf and Ui are the final and initial potential energy of the electron.

(c) For the conservation of energy, the sum of the kinetic energy and potential energy of the electron at the beginning of the motion and at the end must be equal:
$U_i + K_i = U_f + K_f$ (1)
where Ki and Kf are the initial and final kinetic energies.
The electron is initially at rest, so Ki =0, and we can rewrite (1) as
$U_i - U_f = - \Delta U = K_f = \frac{1}{2}m_e v_f^2$
and by using the mass of the electron me, we can find the value of the final velocity of the electron:
$v_f= \sqrt{ -\frac{2 \Delta U}{m_e} }= \sqrt{- \frac{2(-1.9 \cdot 10^{-18} J)}{9.1 \cdot 10^{-31} kg} } =2.04 \cdot 10^6 m/s$

7 0

3 years ago

A cable that weighs 8 lb/ft is used to lift 900 lb of coal up a mine shaft 650 ft deep. Find the work done. Show how to approxim

Harrizon [31]

Answer:

2275000 lb.ft

Explanation:

Let work done on the cable be denoted by: W_ca

Let work done on the coal be denoted by: W_co

Now, dividing the cable into segments, let x represent the length from top of the mine shaft to the segment.

Meanwhile let δx be the length of the segment.

We are told the cable weighs 8 lb/ft. Thus;

Work done on one segment = 8 × δx × x = 8x•δx

Therefore, work done on cable is;

W_ca = ∫8x•δx between the boundaries of 0 and 650

Thus;

W_ca = 4x² between the boundaries of 0 and 650

W_ca = 4(650²) - 4(0²)

W_ca = 1,690,000 lb.ft

Workdone on the 900 lb of coal will be calculated as;

W_co = 900 × 650

W_co = 585000 lb.ft

Thus,

Total work done = W_ca + W_co

Total workdone = 1690000 + 585000

Total workdone = 2275000 lb.ft

6 0

3 years ago

initially, a particle is moving at 5.33 m/s at an angle of 37.9° above the horizontal. Two seconds later, its velocity is 6.11 m

Ainat [17]

Explanation:

Average acceleration is the change in velocity over the change in time:

a = (v − v₀) / t

In the x direction:

aₓ = (6.11 cos (-54.2°) − 5.33 cos (37.9°)) / 2.00

aₓ = -0.316 m/s²

In the y direction:

aᵧ = (6.11 sin (-54.2°) − 5.33 sin (37.9°)) / 2.00

aᵧ = -4.11 m/s²

7 0

3 years ago

Which temperature is lower ? -44°c -55°c

Mice21 [21]

Answer:

-55 degrees c

Explanation:

because negative numbers work in reverse and that makes -55 actually lower.

6 0

2 years ago