All categories

3 years ago

Define various optical properties of engineering materials

Engineering

1 answer:

natta225 [31]3 years ago

3 0

Answer:

The optical properties of matter are studied in optical physics, a subfield of optics. ... Refraction and the material's refraction index.

Explanation:

You might be interested in

You assume the following relationship: u=x1αx2β. You take the following monotonic transformation: v=ln(u). You get a linear rela

ollegr [7]

Answer:

MRS = -2.23 (x2/x1)

Explanation:

The concept of Marginal rate of substitution(MRS) is applied here. MRS occurs when a consumer is willing to give up a good at the expense of another whilst still maintaining the same utility. The principle of partial differentiation is mostly applied in MRS Calculation.

From MRS = -(MUx/MUy)

The steps are as shown in the attachment

7 0

4 years ago

Which statement about direct-mail messages is most accurate? Group of answer choices Direct mail is an effective channel for per

torisob [31]

Answer:

Option A

Direct mail is an effective channel for personalized, tangible, three-dimensional messages that are less invasive than telephone solicitations

Explanation:

In modern days, we've seen email marketing hence option B that "Today's companies no longer send direct-mail messages to market their products or services; they rely exclusively on electronic media instead." These companies rely also on direct mails to their clients. Online marketing often uses direct mail for persuasion of their customers hence the last option is also wrong. Considering that already two statements are identified as inaccurate, the option that "All answer choices are accurate statements about direct-mail messages. " is misleading. Therefore, the only most accurate choice about direct-mail is option A

4 0

4 years ago

An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500

aleksandrvk [35]

Answer:

Exit temperature = 32°C

Explanation:

We are given;

Initial Pressure;P1 = 100 KPa

Cp =1000 J/kg.K = 1 KJ/kg.k

R = 500 J/kg.K = 0.5 Kj/Kg.k

Initial temperature;T1 = 27°C = 273 + 27K = 300 K

volume flow rate;V' = 15 m³/s

W = 130 Kw

Q = 80 Kw

Using ideal gas equation,

PV' = m'RT

Where m' is mass flow rate.

Thus;making m' the subject, we have;

m' = PV'/RT

So at inlet,

m' = P1•V1'/(R•T1)

m' = (100 × 15)/(0.5 × 300)

m' = 10 kg/s

From steady flow energy equation, we know that;

m'•h1 + Q = m'h2 + W

Dividing through by m', we have;

h1 + Q/m' = h2 + W/m'

h = Cp•T

Thus,

Cp•T1 + Q/m' = Cp•T2 + W/m'

Plugging in the relevant values, we have;

(1*300) - (80/10) = (1*T2) - (130/10)

Q and M negative because heat is being lost.

300 - 8 + 13 = T2

T2 = 305 K = 305 - 273 °C = 32 °C

4 0

3 years ago

A 179 ‑turn circular coil of radius 3.95 cm and negligible resistance is immersed in a uniform magnetic field that is perpendicu

suter [353]

Answer:

The energy, that is dissipated in the resistor during this time interval is 153.6 mJ

Explanation:

Given;

number of turns, N = 179

radius of the circular coil, r = 3.95 cm = 0.0395 m

resistance, R = 10.1 Ω

time, t = 0.163 s

magnetic field strength, B = 0.573 T

Induced emf is given as;

$emf= N\frac{d \phi}{dt}$

where;

ΔФ is change in magnetic flux

ΔФ = BA = B x πr²

ΔФ = 0.573 x π(0.0395)² = 0.002809 T.m²

$emf = N\frac{d \phi}{dt} = 179(\frac{0.002809}{0.163} ) = 3.0848 \ V$

According to ohm's law;

V = IR

I = V / R

I = 3.0848 / 10.1

I = 0.3054 A

Energy = I²Rt

Energy = (0.3054)² x 10.1 x 0.163

Energy = 0.1536 J

Energy = 153.6 mJ

Therefore, the energy, that is dissipated in the resistor during this time interval is 153.6 mJ

6 0

3 years ago

Question 8 (8 points)

ivanzaharov [21]

My stick is 12 in on my ruler so ik I’m good idk bout you tip

7 0

3 years ago

Read 2 more answers