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4 years ago

How can you tell if a diamond real or fake

Physics

2 answers:

rewona [7]4 years ago

8 0

Scratch it on a peice of glass

notsponge [240]4 years ago

6 0

You can tell if a diamond is real or fake by checking its edges. Real diamonds have sharp edges, while fake ones don't. Another way is to bring it to a jewelry shop, or have a professional look at it, which is the best and easiest way. You can also see if a diamond is real or not by placing it on top of a newspaper, and to see if you can read the text through the diamond. If you can, then it is a fake, because real diamonds contort light, making the letters of the newspaper blurry.

~Hope I helped!~

You might be interested in

When you drop an object from a certain height, it takes time T to reach the ground with no air resistance. If you dropped it fro

alexandr402 [8]

Answer:

Explanation:

Given

When we drop an object from height , suppose h

it takes time T

using equation of motion

$h=ut+\frac{1}{2}at^2$

where

$h=displacement$

$u=initial\ velocity$

$a=acceleration$

$t=time$

here $u=0$ because it dropped from a certain height

$h=\frac{1}{2}gT^2$

$T=\sqrt{\frac{2h}{g}}$

When height is increases to three times of original height

i.e. $h'=3 h$

then time period becomes

$T'=\sqrt{\frac{2\times 3h}{g}}$

$T'=\sqrt{3}T$

5 0

3 years ago

You’ve had practice calculating the grams of hydrogen gas, but it is also possible to calculate the amount of oxygen gas produce

alekssr [168]

Answer:41.991ml

Explanation:

Equations: 2 H2O → 4H+ + 4e + O2 OXIDATION

2 H+ + 2e → H2 REDUCTION

Electrolysis is the chemical decomposition of compounds when electricity is made to pass through a molten compound or solution.

from the oxidation reaction:

1moles of oxygen requires 4moles of electrons to be discharged at the product

F=96500C/mol

Quantity of charge Q=It

=60*60*0.201A

Q=723.6C

Mole=Q/(F*mole ratio of electron)

Mole= 723.6/(4*96500)

Mole=((1809)/(965000))

M=0.0018746114

M1/M2=V1/V2

1/0.00187=22.4dm^3/V2

V2=22.4*0.00187

V2=0.04199129534dm^3

41.99129534ml

5 0

3 years ago

The speed of sound in room temperature (20°C) air is 343 m/s; in room temperature helium, it is 1010 m/s. The fundamental freque

Lera25 [3.4K]

Answer: f = 927.55Hz

Explanation: Since the the tube is open-closed, the length of air and the wavelength of sound passing through the tube is given below

L = λ/4 where λ = wavelength.

speed of sound in air = v = 343m/s.

fundamental frequency of open closed tube = 315Hz

λ = 4L.

v = fλ

343 = 315 * 4L

343 = 1260 * L

L = 343/ 1260

L = 0.27m

In the same tube of length L = 0.27m but different medium ( helium), the speed of sound is 1010m/s.

The length of tube and wavelength are related by the formulae below

L = λ/4, λ=4L

λ = 4 * 0.27

λ = 1.087m.

v = fλ

1010 = f * 1.087

f = 1010/1.807

f = 927.55Hz

4 0

3 years ago

Which cycle causes earthquake and volcanic activity? carbon water tectonic phosphorus

erik [133]

The answer would be tectonic. hope this helps (:

6 0

3 years ago

Read 2 more answers

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at

denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m _____ eqn (1)

FOR HYDROGEN:

v = √(3KT)/m = 2000 m/s _____ eqn (2)

FOR OXYGEN:

velocity of oxygen = √(3KT)/(mass of oxygen)

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

velocity of oxygen = (1/4)(2000 m/s) = 500 m/s

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

K.E of Oxygen = 4000 J

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

v = 2149.24 m/s

8 0

4 years ago