Ask question

Ask question

All categories

1 month ago

12

A permeability test was run on a compacted sample of dirty sandy gravel. The sample was 175 mm long and the diameter of the mold

175 mm. In 90 s the discharge under a constant head of 38 em was 405 cm3 The sample had a dry mass of 4950 g and its Ps was 2710 kg!m3 Calculate (a) the coefficient of permeability, (b) the seepage velocity, and (c) the discharge velocity during the test.

1 answer:

Maru [3.3K]1 month ago

8 0

Answer:

(a). The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

Explanation:

Given that,

Length = 175 mm

Diameter = 175 mm

Time = 90 sec

Volume= 405 cm³

We need to calculate the discharge

Using formula of discharge

$Q=\dfrac{V}{t}$

Insert the values into the formula

$Q=\dfrac{405}{90}$

$Q=4.5\ cm^3/s$

(a). We will compute the coefficient of permeability

Applying the formula for permeability

$Q=kiA$

$k=\dfrac{Q}{iA}$

$k=\dfrac{Ql}{Ah}$

Where, Q=discharge

l = length

A = cross-sectional area

h=constant head that causes flow

Plugging the value into the formula

$k=\dfrac{4.5\times175\times10^{-1}}{\dfrac{\pi(175\times10^{-1})^2}{4}\times38}$

$k=8.6\times10^{-3}\ cm/s$

The coefficient of permeability measures as $8.6\times10^{-3}\ cm/s$ .

(c). To ascertain the discharge velocity during the testing phase

Utilizing the discharge velocity formula

$v=ki$

$v=\dfrac{kh}{l}$

Substituting the values into the equation

$v=\dfrac{8.6\times10^{-3}\times38}{17.5}$

$v=0.0187\ cm/s$

The discharge velocity during the test measures 0.0187 cm/s.

Thus, (a). The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(b). The seepage velocity computes at 0.0330 cm/s.

(c). The observed discharge velocity during the test equals 0.0187 cm/s.

You might be interested in

Potential energy matter has a result of its ____ or ____.

Yuliya22 [3333]

Position or composition

3 0

2 months ago

Read 2 more answers

An object of mass 8.0 kg is attached to an ideal massless spring and allowed to hang in the Earth's gravitational field. The spr

inna [3103]

The derived frequency equals 2.63 Hz. Explanation: For an object weighing 8.0 kg with a spring stretching 3.6 cm, calculations involving the spring constant and oscillation frequency lead to this specific oscillation rate.

4 0

1 month ago

Read 2 more answers

Derive an equation for the acceleration of block 3 for any arbitrary values of m3 and m2. Express your answer in terms of m3, m2

Ostrovityanka [3204]

The complete question is;

Block 1 sits on the floor with block 2 resting atop it. Block 3, which is stationary on a frictionless table, is attached to block 2 via a string that passes over a pulley depicted in the illustration below. Both the string and pulley have negligible mass.

Once block 1 is taken away without impacting block 2.

Derive an equation for the acceleration of block 3 considering arbitrary values for m3 and m2. Express your answer in terms of m3, m2, and relevant physical constants as needed.

Answer:

a = (m2)g/(m3 + m2)

Explanation:

Examining the attached illustration, by analyzing the free body diagram for block 3 and utilizing Newton's first law of motion, we reach the following formula;

T = (m3)a - - - (eq 1)

where;

T is the tension in the string

a is acceleration

m3 is the mass of block 3

Simultaneously, doing the same for Block 2, the free body diagram yields the equation; (m2)g - T = (m2)a

Rearranging for T results in;

T = (m2)g - (m2)a - - - (eq 2)

where;

g represents acceleration due to gravity

T is the tension in the string

a is acceleration

m2 is the mass of block 2

To deduce the acceleration, we will substitute (m3)a in place of T in eq 2.

Thus;

(m3)a = (m2)g - (m2)a

(m3)a + (m2)a = (m2)g

a(m3 + m2) = (m2)g

a = (m2)g/(m3 + m2)

3 0

28 days ago

If Pete ( mass=90.0kg) weights himself and finds that he weighs 30.0 pounds, how far away from the surface of the earth is he

serg [3582]

The answer is 9938.8 km. Explanation: 1 pound-force = 4.48 N. Hence, 30.0 pounds-force = 134.4 N. The gravitational force between Earth and an object on its surface is defined by: Where M denotes Earth’s mass, m is the object's mass, and R represents the Earth's radius (6371 km). To determine height (h) above Earth's surface, we compare ratios. Ultimately, Pete's weight would be 30 pounds at a height of 9938.8 km from the Earth's surface.

5 0

1 month ago

One consequence of Einstein's theory of special relativity is that mass is a form of energy. This mass-energy relationship is pe

ValentinkaMS [3465]

Answer:

The energy unit is expressed as kg-m/s or Joules.

Explanation:

The relationship between mass and energy in physics is represented by:

$E=mc^2$

Where

m denotes the mass of the object

c signifies the speed of light

In the SI system, mass is measured in kilograms and the speed of light in m/s. Therefore, energy is defined in kg-m/s, which is equal to Joules.

Thus, the appropriate SI unit for energy is kg-m/s or Joules. This concludes the explanation.

6 0

2 months ago

Read 2 more answers