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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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Tamara has received multiple bids from general contractors to install a new pool in her backyard. How should she go about select
vovikov84 [41]

Answer:

Option C

Explanation:

Tamara must choose the contractor who can provide best services among all and also has a fair pricing. Hence, option 3 is correct

If she chooses the contractor with low price, there are chances that the quality will not be good. Hence, option 2 is incorrect

Option 4 is incorrect because the contractor must take inputs of the contracting person or agency.

Option 1 is incorrect because selecting some one on the basis of reviews can be a biased selection of contractor.

7 0
3 years ago
In a reversible process both the system and surrondings can be returned to their initial states. a)-True b)-False
Nuetrik [128]

Answer:

a)-True

Explanation:

When the direction of a process is reversed through infinitesimal changes to the system. The infinitesimal changes are done by the surroundings. After the process has been reversed the system and surroundings go back to their original state.

The changes take an infinite amount of time so, it is not possible practically.

6 0
3 years ago
An offshore oil rig will drill into the ocean floor which is located 1 mile below the water surface. Determine the pressure at t
Luden [163]

Answer:

15.65 MPa 150 times of human body internal pressure

Explanation:

Given:

- Depth of the ocean for offshore drilling d = 1 mile

Find:

Determine the pressure at that location

What would happen to an unprotected, exposed person at that depth?

Solution:

- The pressure at a certain depth of a fluid can be calculated with:

                                       P = p_w*g*d

Where, P is the pressure , p_w is the density of water ( 997 kg / m^3 ).

- Hence @ d = 1.0 mile = 1.6 km = 1600 m:

                                       P = 997*9.81*1600

                                       P = 15.65 MPa

- Whereas the pressure inside a human body is 101 KPa, Pressure under ocean @  1 mile of depth is 150 times in magnitude, enough to crush the human body!

4 0
3 years ago
In a mechanical assembly operation, the first work unit required 7.83 min to complete and the learning rate for mechanical assem
Marianna [84]

Answer:

(a) 6.5774 minutes

(b) 4.3879 minutes

(c) 2.459 minutes

(d) 245.902 minutes

Explanation:

The learning curve model follows the following equation:

Y=aX^b

where Y are the units time to produce X units, and a is the time for assembly the first unit. Additionally, b is calculated as:

b=\frac{ln(r)}{ln(2)}

Where r is the learning rate for mechanical assembly. So, b is equal to:

b=\frac{ln(0.84)}{ln(2)}=-0.2515

Then, the equation is:

Y=7.83X^{-0.2515}

Finally, the unit times to produce the second unit are:

Y=7.83(2)^{-0.2515}=6.5774

The 10th unit:

Y=7.83(10)^{-0.2515}=4.3879

The 100th unit:

Y=7.83(100)^{-0.2515}=2.459

Then, the total cumulative time T to produce 100 units in the Crawford model is calculated as:

T=7.83\frac{(100.5)^{1-0.2515}-(0.5)^{1-0.2515})}{1-0.2515} \\T=323.5383

7 0
3 years ago
Using a forked rod, a 0.5-kg smooth peg P is forced to move along the vertical slotted path r = (0.5 θ) m, whereθ is in radians.
-BARSIC- [3]

Answer:

N_c = 3.03 N

F = 1.81 N

Explanation:

Given:

- The attachment missing from the question is given:

- The given expressions for the radial and θ direction of motion:

                                       r = 0.5*θ

                                       θ = 0.5*t^2              ...... (correction for the question)

- Mass of peg m = 0.5 kg

Find:

a) Determine the magnitude of the force of the rod on the peg at the instant t = 2 s.

b) Determine the magnitude of the normal force of the slot on the peg.

Solution:

- Determine the expressions for radial kinematics:

                                        dr/dt = 0.5*dθ/dt

                                        d^2r/dt^2 = 0.5*d^2θ/dt^2

- Similarly the expressions for θ direction kinematics:

                                        dθ/dt = t

                                        d^2θ/dt^2 = 1

- Evaluate each at time t = 2 s.

                                        θ = 0.5*t^2 = 0.5*2^2 = 2 rad -----> 114.59°

                                        r = 1 m , dr / dt = 1 m/s , d^2 r / dt^2 = 0.5 m/s^2

- Evaluate the angle ψ between radial and horizontal direction:

                                        tan Ψ = r / (dr/dθ) = 1 / 0.5

                                        Ψ = 63.43°

- Develop a free body diagram (attached) and the compute the radial and θ acceleration:

                                        a_r = d^2r / dt^2 - r * dθ/dt

                                        a_r = 0.5 - 1*(2)^2 = -3.5 m/s^2

                                        a_θ =  r * (d^2θ/dt^2) + 2 * (dr/dt) * (dθ/dt)

                                        a_θ = 1(1) + 2*(1)*(2) = 5 m/s^2

- Using Newton's Second Law of motion to construct equations in both radial and θ directions as follows:

Radial direction:              N_c * cos(26.57) - W*cos(24.59) = m*a_r

θ direction:                      F  - N_c * sin(26.57) + W*sin(24.59) = m*a_θ

Where, F is the force on the peg by rod and N_c is the normal force on peg by the slot. W is the weight of the peg. Using radial equation:

                                       N_c * cos(26.57) - 4.905*cos(24.59) = 0.5*-3.5

                                       N_c = 3.03 N

                                       F  - 3.03 * sin(26.57) + 4.905*sin(24.59) = 0.5*5

                                       F = 1.81 N

4 0
3 years ago
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