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3 years ago

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Calculate the time required (in years) for water to penetrate a layer of clay that is 40 cm deep when exposed to a hydraulic gra

dient of 1 cm/cm. The permeability of clay is 1x10-8 cm/sec. If there is 30 cm static head of water on the clay layer, how long will it take for moisture to penetrate the 40 cm clay layer (in years)?

2 answers:

guapka [62]3 years ago

5 0

Answer:

The time required to penetrate the 40 cm clay layer is 126.82 years

Explanation:

Given:

Hydraulic gradient = 1 $\frac{cm}{cm}$

Permeability of clay $K = 1 \times 10^{-8}$ $\frac{cm}{sec}$

Time required to penetrate 40 cm clay layer,

$= \frac{40}{K}$

$= \frac{40}{1 \times 10^{-8} }$

$= 40 \times 10^{8}$ sec

But we have to find time in years,

$= \frac{40 \times 10^{8} }{24 \times 60 \times 60 \times 365}$ years

$= 126.82$ years

Therefore, the time required to penetrate the 40 cm clay layer is 126.82 years

Free_Kalibri [48]3 years ago

3 0

Answer:

its a

Explanation:

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A squirrel falls from this tree after being shocked by the falling apples. If the

Alchen [17]

Answer:

0.85m

Explanation:

Given parameters:

Height of fall = 3.5m

Unknown:

Duration of fall = ?

Solution:

To solve this problem, we apply the right motion equation.

Since we know the height, we can use the equation below;

S = ut + $\frac{1}{2} gt^{2}$

S is the height

u is the initial velocity = 0m/s

t is the time

g is the acceleration due to gravity

3.5 = 0 + $\frac{1}{2}$ x 9.8 x t²

3.5 = 4.9t²

t² = $\frac{3.5}{4.9}$

t² = 0.71

t = √0.71 = 0.85m

3 0

3 years ago

A square coil of wire with 15 turns and an area of 0.40 m2 is placed parallel to a magnetic field of 0.75 T. The coil is flipped

Drupady [299]

Answer:

The magnitude of the average induced emf is 90V

Explanation:

Given;

area of the square coil, A = 0.4 m²

number of turns, N = 15 turns

magnitude of the magnetic field, B = 0.75 T

time of change of magnetic field, t = 0.05 s

The magnitude of the average induced emf is given by;

E = -NAB/t

E = -(15 x 0.4 x 0.75) / 0.05

E = -90 V

|E| = 90 V

Therefore, the magnitude of the average induced emf is 90V

6 0

3 years ago

A. According to theory, the period T of a simple pendulum is T = 2????√ ???? ???? a. If ???? is measured as ???? = 1.40 ± 0.01 m

salantis [7]

Answer:

a) T = (2,375 ± 0.008) s , b) When comparing this interval with the experimental value we see that it is within the possible theoretical values.

Explanation:

a) The period of a simple pendulum is

T = 2π √ L / g

Let's calculate

T = 2π √1.40 / 9.8

T = 2.3748 s

The uncertainty of the period is

ΔT = dT / dL ΔL

ΔT = 2π ½ √g/L 1/g ΔL

ΔT = π/g √g/L ΔL

ΔT = π/9.8 √9.8/1.4 0.01

ΔT = 0.008 s

The result for the period is

T = (2,375 ± 0.008) s

b) the experimental measure was T = 2.39 s ± 0.01 s

The theoretical value is comprised in a range of [2,367, 2,387] when we approximate this measure according to the significant figures the interval remains [2,37, 2,39].

When comparing this interval with the experimental value we see that it is within the possible theoretical values.

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4 years ago

Which of the following electrical components is a temporary electrical energy storage device?

7nadin3 [17]

Answer:

A capacitor

Explanation:

Because it can store electric energy when disconnected from its charging circuit. Commonly used in electronic devices to maintain power supply while batteries change.

Hope this helps! :)

6 0

3 years ago

As you know, a common example of a harmonic oscillator is a mass attached to a spring. In this problem, we will consider a horiz

Eddi Din [679]

a)E= U + K = $\frac{1}{2}$ kx² + $\frac{1}{2}$ mv²

The total energy of the system at any point in the motion is equal to the sum of the elastic potential energy of the spring, U, and of the kinetic energy of the mass, K:

E= U + K = $\frac{1}{2}$ kx² + $\frac{1}{2}$ mv²

where

'k' represents the spring constant

'x' is the compression/stretching of the spring with respect to its equilibrium position

'm' is the mass of the block attached to the spring

and 'v' is the speed of the block

b) <em>A=</em> $\sqrt{\frac{2E}{k}}$ <em> </em>

The amplitude of the motion compares to the most extreme displacement of the mass-spring system. The displacement of the system, x(t), at time t, for a simple harmonic oscillator is given by,

x= Asin(ωt+∅)

where

amplitude is 'A'

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency of the motion

t is the time

$\phi$ is the phase (we can take $\phi$ =0 )

The amplitude of the motion occurs when the displacement of the motion is maximum: x=A. Regarding energy, the mass-spring system is at its maximum displacement (x=A) when all the mechanical energy of the framework is elastic potential energy, so when the kinetic energy is zero:

K= $\frac{1}{2}mv^2$ =0

E= $\frac{1}{2}kA^2\\$ -->(1)

<em>A=</em> $\sqrt{\frac{2E}{k}}$ <em> </em>

c) $v_{max}=\omega A$ <u></u>

When the elastic potential energy is zero, the maximum speed of the system occurs i.e U=0 and the kinetic energy is maximum, so:

U=0

E= $\frac{1}{2}mv_{max}^2$

According to the law of conservation of the mechanical energy, this energy must be equal to the energy of the system at its maximum displacement (1), so we can write

$\frac{1}{2}kA^2=\frac{1}{2}mv_{max}^2$

and solving for $v_{max}$ we find an expression for the maximum speed:

$v_{max}=\sqrt{\frac{kA^2}{m}}=\sqrt{\frac{k}{m}}A=\omega A$

<h2><u></u> $v_{max}=\omega A$ <u></u></h2>

4 0

3 years ago