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KATRIN_1 [288]
3 years ago
6

What is the difference between a natural and artificial diamond ​

Engineering
2 answers:
IgorC [24]3 years ago
7 0

Answer:

So, if natural and synthetic diamonds have the same makeup, are there any differences?

Synthetic diamonds do differ from natural stones, but not as far as their fundamental structure is concerned. The main difference that can be observed is in the clarity of the stones.

Whereas most natural diamonds have internal flaws that occurred during the stones’ formation, synthetics tend to be cleaner.

This is because they are created through a controlled process designed to minimize defects, and as a result, there are fewer flaws in the crystal structure of synthetics

\

muminat3 years ago
3 0

Answer:

Lab-grown diamonds are created in a laboratory, often produced in just a matter of weeks. There is a tiny chemical difference between the two, as natural diamonds often contain a very small amount of nitrogen, while synthetic diamonds do not.Nov 23, 2020

Explanation:

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There is an electric field near the Earth's surface whose magnitude is about 145 V/m . How much energy is stored per cubic meter
weqwewe [10]

Answer:

u_e = 9.3 * 10^-8 J / m^3  ( 2 sig. fig)

Explanation:

Given:

- Electric Field strength near earth's surface E = 145 V / m

- permittivity of free space (electric constant) e_o =  8.854 *10^-12 s^4 A^2 / m^3 kg

Find:

- How much energy is stored per cubic meter in this field?

Solution:

- The solution requires the energy density stored between earth's surface and the source of electric field strength. The formula for charge density is given by:

                                        u_e = 0.5*e_o * E^2

- Plug in the values given:

                                        u_e = 0.5*8.854 *10^-12 *145^2

                                        u_e = 9.30777 * 10^-8  J/m^3

5 0
3 years ago
As the junior engineer at the Mesabi Range Hydraulic Engineering Company located in Ely, Minnesota, you have been tasked with de
katen-ka-za [31]

yes it will

Explanation:

5 0
3 years ago
The electron concentration in silicon at T = 300 K is given by
puteri [66]

Answer:

E=1.44*10^-7-2.6exp(\frac{-x}{18} )v/m

Explanation:

From the question we are told that:

Temperature of silicon T=300k

Electron concentration n(x)=10^{16}\exp (\frac{-x}{18})

                                        \frac{dn}{dx}=(10^{16} *(\frac{-1}{16})\exp\frac{-x}{16})

Electron diffusion coefficient is Dn = 25cm^2/s \approx 2.5*10^{-3}

Electron mobility is \mu n = 960 cm^2/V-s \approx0.096m/V

Electron current density Jn = -40 A/cm^2 \approx -40*10^{4}A/m^2

Generally the equation for the semiconductor is mathematically given by

Jn=qb_n\frac{dn}{dx}+nq \mu E

Therefore

-40*10^{4}=1.6*10^{-19} *(2.5*10^{-3})*(10^{16} *(\frac{-1}{16})\exp\frac{-x}{16})+(10^{16}\exp (\frac{-x}{18}))*1.6*10^{-19}*0.096* E

E=\frac{-2.5*10^-^7 exp(\frac{-x}{18})+40*10^{4}}{1.536*10^-4exp(\frac{-x}{18} )}

E=1.44*10^-7-2.6exp(\frac{-x}{18} )v/m

7 0
3 years ago
Calculate the rate at which body heat is conducted through the clothing of a skier in a steady- state process, given the followi
olga2289 [7]

Answer:

230.4W

Explanation:

Heat transfer by conduction consists of the transport of energy in the form of heat through solids, in this case a jacket.

the equation is as follows

Q=\frac{KA(T2-T1)}{L} \\

Where

Q=heat

k=conductivity=0.04

A=Area=1.8m^2

T2=33C

T1=1C

L=thickness=1cm=0.01mQ=\frac{(0.04)(1.8m^2)(33-1)}{0.01m}

Q=230.4W

the skier loses heat at the rate of 230.4W

4 0
3 years ago
If a heat engine has an efficiency of 30% and its power output is 600 W, what is the rate of heat input from the combustion phas
jarptica [38.1K]

Answer:

The heat input from the combustion phase is 2000 watts.

Explanation:

The energy efficiency of the heat engine (\eta), no unit, is defined by this formula:

\eta = \frac{\dot W}{\dot Q} (1)

Where:

\dot Q - Heat input, in watts.

\dot W - Power output, in watts.

If we know that \dot W = 600\,W and \eta = 0.3, then the heat input from the combustion phase is:

\eta = \frac{\dot W}{\dot Q}

\dot Q = \frac{\dot W}{\eta}

\dot Q = \frac{600\,W}{0.3}

\dot Q = 2000\,W

The heat input from the combustion phase is 2000 watts.

8 0
3 years ago
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