A 0.69 kg rubber ball has a speed of 1.4 m/s at point a and kinetic energy 7.5 j at point

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

LUCKY_DIMON [66]

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

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

Put the value into the formula

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

$Q=4.5\ cm^3/s$

(a). We need to calculate the coefficient of permeability

Using formula of coefficient of permeability

$Q=kiA$

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

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

Where, Q=discharge

l = length

A = cross section area

h=constant head causing flow

Put 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 is $8.6\times10^{-3}\ cm/s$ .

(c). We need to calculate the discharge velocity during the test

Using formula of discharge velocity

$v=ki$

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

Put the value into the formula

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

$v=0.0187\ cm/s$

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

(b). We need to calculate the volume of solid in the ample

Using formula of volume

$V_{s}=\dfrac{M_{s}}{V_{s}}$

Put the value into the formula

$V_{s}=\dfrac{4950\times10^{-3}}{2710}$

$V_{s}=1826.56\ cm^3$

We need to calculate the volume of the soil specimen

Using formula of volume

$V=A\times L$

Put the value into the formula

$V=\dfrac{\pi(17.5)^2}{4}\times17.5$

$V=4209.24\ cm^3$

We need to calculate the volume of the voids

$V_{v}=V-V_{s}$

Put the value into the formula

$V_{v}=4209.24-1826.56$

$V_{v}=2382.68\ cm^3$

We need to calculate the seepage velocity

Using formula of velocity

$Av=A_{v}v_{s}$

$v_{s}=\dfrac{Av}{A_{v}}$

$v_{s}=\dfrac{V}{V_{v}}\times v$

Put the value into the formula

$v_{s}=\dfrac{4209.24}{2382.68}\times0.0187$

$v_{s}=0.0330\ cm/s$

The seepage velocity is 0.0330 cm/s.

Hence, (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.

8 0

4 years ago

Directions: Select the choice that best fits each statement. The following question(s) refer to the following concepts related t

borishaifa [10]

Answer:

(E) Second law of thermodynamics

Explanation:

The second law of thermodynamics can be understood according to Clausius' words: In an isolated system, no process can occur if a decrease in the total entropy of the system is associated with it. These processes are associated with energy transformations, in which a variable is introduced, called entropy that indicates the notion of disorder. Therefore, in any isolated process, the disorder can only grow.

4 0

3 years ago

Read 2 more answers

A train is moving in the positive direction down a track. First the train speeds up, and then it slows down. What is its acceler

tatuchka [14]

C.

The train first sped up, giving it a positive acceleration in the beginning. This eliminates D since that choice states that it begins with a negative acceleration. This also eliminates B since that choice states that the train only had a negation acceleration.

Next, the train slows down, giving it a negative acceleration. We’re looking for the answer choice that starts with a positive acceleration and ends with a negative one. This makes C the correct answer. Hope this helps!

8 0

3 years ago

Read 2 more answers

Why is hydroelectric power considered to be an example of multiple transfers of energy?

Radda [10]

Hydroelectric power is considered to be an example of multiple transfers of energy because potential energy when the water is in the reservoir turns into kinetic energy making the water move to the dam and then the dam moves into the turbine turning it to electric energy.

7 0

3 years ago

Which is an example of a mixture made of parts that can easily be separated?

Gemiola [76]

Answer

sand and water is an example of a mixture made of parts that can easily be separated

Explatnation

mixture is a material which is formed by mixing two or more different types of materials. No chemical reaction occurs.During combination of two or more type of material to form a mixture.

Example of mixture are

oil and water

sand and water

8 0

3 years ago