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2 years ago

Military glorification through mosaics, relief carvings and triumphant arches are often associated with?

1 answer:

8 0

Military glorification through mosaics, relief carvings and triumphant arches are often associated with roman architectural monument and is the correct choice.

<h3>What is a Monument?</h3>

This can be defined as a structure which is usually large and is used to commemorate the history of an influential person. This helps to serve as an example and also a reminder as to why the performance of good deeds are important

The use of military glorification through mosaics, relief carvings and triumphant arches were also often used by the Romans in other to commemorate war victories and the succession of a new ruler which is known as the emperor.

Read more about Roman monuments here brainly.com/question/15786258

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If you had a match and a lantern and a candle in the dark which one would you choose to light.

PSYCHO15rus [73]

Answer:

The match

Explanation:

You can light both the lantern and the candle if you light the match first.

I don't know of this is a homework question, but I answered it anyway :)

5 0

3 years ago

Read 2 more answers

Given asphalt content test data:

VMariaS [17]

Answer:

hello your question is incomplete attached below is the complete question

A) overall mean = 5.535, standard deviation ≈ 0.3239

B ) upper limit = 5.85, lower limit = 5.0

C) Not all the samples meet the contract specifications

D) fluctuation ( unstable Asphalt content )

Explanation:

B) The daily average asphalt content has to obtained in order to determine the upper and lower control limits using an average asphalt content of 5.5% +/- 0.5% everyday

The upper limit : 14 may = ( 5.8 + 5.1 ) / 2 = 5.85

The lower limit : 16 may = ( 5.2 + 4.8 ) / 2 = 5.0

attached below is the required plot

C ) Not all the samples meet the contract specifications and the samples that do not meet up are samples from :

15 may and 16 may . this is because their Asphalt contents are 6.2 and 4.8 respectively and sample number 18 and 20

D ) what can be observed is that the ASPHALT content fluctuates between the dates while the contract specification is fixed

5 0

4 years ago

A smooth sphere with a diameter of 6 inches and a density of 493 lbm/ft^3 falls at terminal speed through sea water (S.G.=1.0027

Pachacha [2.7K]

Given:

diameter of sphere, d = 6 inches

radius of sphere, r = $\frac{d}{2}$ = 3 inches

density, $\rho}$ = 493 lbm/ $ft^{3}$

S.G = 1.0027

g = 9.8 m/ $m^{2}$ = 386.22 inch/ $s^{2}$

Solution:

Using the formula for terminal velocity,

$v_{T}$ = $\sqrt{\frac{2V\rho g}{A \rho C_{d}}}$ (1)

$(Since, m = V\times \rho)$

where,

V = volume of sphere

$C_{d}$ = drag coefficient

Now,

Surface area of sphere, A = $4\pi r^{2}$

Volume of sphere, V = $\frac{4}{3} \pi r^{3}$

Using the above formulae in eqn (1):

$v_{T}$ = $\sqrt{\frac{2\times \frac{4}{3} \pir^{3}\rho g}{4\pi r^{2} \rho C_{d}}}$

$v_{T}$ = $\sqrt{\frac{2gr}{3C_{d}}}$

$v_{T}$ = $\sqrt{\frac{2\times 386.22\times 3}{3C_{d}}}$

Therefore, terminal velcity is given by:

$v_{T}$ = $\frac{27.79}{\sqrt{C_d}}$ inch/sec

3 0

3 years ago

Question 1: Final Results = What are the values of the resistances such that the gain = -100, Rin = 1 MI2. Don't use resistances

lidiya [134]

Answer:

Explanation:

In a study of algebra, you will encounter many families of equations, or groups of

equations that share common characteristics. Of interest to us here is the family of

linear equations in one variable, a study that lays the foundation for understanding

more advanced families. In addition to solving linear equations, we’ll use the skills we

develop to solve for a specified variable in a formula, a practice widely used in science,

business, industry, and research.

A. Solving Linear Equations Using Properties of Equality

An equation is a statement that two expressions are

equal. From the expressions and

we can form the equation

which is a linear equation in one variable. To solve

an equation, we attempt to find a specific input or xvalue that will make the equation true, meaning the

left-hand expression will be equal to the right. Using

Table 1.1, we find that is a

true equation when x is replaced by 2, and is a false

equation otherwise. Replacement values that make

the equation true are called solutions or roots of the equation.

4 0

2 years ago

An automated transfer line is to be designed. Based on previous experience, the average downtime per occurrence = 5.0 min, and t

IRINA_888 [86]

Answer:

a) 28 stations

b) Rp = 21.43

E = 0.5

Explanation:

Given:

Average downtime per occurrence = 5.0 min

Probability that leads to downtime, d= 0.01

Total work time, Tc = 39.2 min

a) For the optimum number of stations on the line that will maximize production rate.

Maximizing Rp =minimizing Tp

Tp = Tc + Ftd

$= \frac{39.2}{n} + (n * 0.01 * 5.0)$

$= \frac{39.2}{n} + (n * 0.05)$

At minimum pt. = 0, we have:

dTp/dn = 0

$= \frac{-39.2}{n^2} + 0.05 = 0$

Solving for n²:

$n^2 = \frac{39.2}{0.05} = 784$

$n = \sqrt{784} = 28$

The optimum number of stations on the line that will maximize production rate is 28 stations.

b) $Tp = \frac{39.2}{28} + (28 * 0.01 * 5)$

Tp = 1.4 +1.4 = 2.8

The production rate, Rp =

$\frac{60min}{2.8} = 21.43$

The proportion uptime,

$E = \frac{1.4}{2.8} = 0.5$

3 0

3 years ago