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3 years ago

A student creates interest in a visualization by contrasting the amount of light and dark. what color element is used?

1 answer:

3 0

A student creates interest in a visualization by contrasting the amount of light and dark. What color element is used?Grayscale

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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

3 years ago

What's the answers for all of the questions ?///

Maru [420]

More energy, colour and I think less bright

8 0

3 years ago

Would the frequency of the angular simple harmonic motion (SHM) of the balance wheel increase or decrease if the dimensions of t

storchak [24]

Answer:

Yes the frequency of the angular simple harmonic motion (SHM) of the balance wheel increases three times if the dimensions of the balance wheel reduced to one-third of original dimensions.

Explanation:

Considering the complete question attached in figure below.

Time period for balance wheel is:

$T=2\pi\sqrt{\frac{I}{K}}$

$I=mR^{2}$

m = mass of balance wheel

R = radius of balance wheel.

Angular frequency is related to Time period as:

$\omega=\frac{2\pi}{T}\\\omega=\sqrt{\frac{K}{I}} \\\omega=\sqrt{\frac{K}{mR^{2}}$

As dimensions of new balance wheel are one-third of their original values

$R_{new}=\frac{R}{3}$

$\omega_{new}=\sqrt{\frac{K}{mR_{new}^{2}}}\\\\\omega_{new}=\sqrt{\frac{K}{m(\frac{R}{3})^{2}}}\\\\\omega_{new}={3}\sqrt{\frac{K}{mR^{2}}}\\\\\omega_{new}={3}\omega$

5 0

3 years ago

I will pick the most Brainliest but... i need u to tell me how too

katovenus [111]

Answer:

I think it’s a

Explanation:

8 0

3 years ago

Read 2 more answers

An object that is dropped straight down from a height of 100 m has a vertical change in position that is less than that of an id

sergey [27]

Objects dropped straight or thrown horizontally from the same height
change their vertical velocity at the same rate, and fall through equal
vertical distances in equal time intervals.

The statement is false.

5 0

3 years ago