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evablogger [386]
3 years ago
10

Pyrite is called "fool's gold" because it's looks a lot like gold. Which properties can be used to tell gold and pyrite apart?

Physics
2 answers:
AlladinOne [14]3 years ago
7 0
Density, streak, and geochemical signature through use of inductively coupled plasma mass spectrometry or Fourier transform infrared spectroscopy
umka2103 [35]3 years ago
4 0

Answer:

On the basis of physical and chemical properties, pyrite can be differentiated from real gold.

Explanation:

Some properties which can be used to differentiate pyrite from real gold are given below:

i) Real gold is made of same kind of atoms while pyrite is made of combinations of atoms like Iron and Sulphur.

ii) Pyrite is lighter in color than real gold.

iii) When we heat pyrite and real gold, both show different colors at higher temperatures.  

iv) Pyrite is less dense and harder than real gold.

v) Real gold has weight 1.5 times than that of pyrite.

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A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path wit
makvit [3.9K]

Answer:

Final velocity of the block = 2.40 m/s east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Mass of bullet = 0.0140 kg

Consider east as positive.

Initial velocity of bullet = 205 m/s

Mass of Block = 1.8 kg

Initial velocity of block = 0 m/s

Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

Final velocity of bullet = -103 m/s

We need to find final velocity of the block( u )

Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u

We have

            2.87 = -1.442 + 1.8 u

               u = 2.40 m/s

Final velocity of the block = 2.40 m/s east.

7 0
4 years ago
The SI unit of power is???​
attashe74 [19]

Answer:

Watts or Wt's

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3 years ago
A 20-Kg child is on a swing attached to 3.0 m-long chains. The child swings back and forth, swinging out to a 60-degree angle. (
kvv77 [185]

Answer:

 v = 29.4 m / s

Explanation:

For this exercise we can use the conservation of mechanical energy

Lowest starting point.

          Em₀ = K = ½ m v²

final point. Higher

          Em_{f} = U = m g h

Let's use trigonometry to lock her up

          cos 60 = y / L

          y = L cos 60

Height is the initial length minus the length at the maximum angle

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           h = L (1- cos 60)

energy is conserved

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          ½ m v² = mgL (1 - cos 60)

         v = 2g L (1- cos 60)

 

let's calculate

          v² = 2 9.8 3.0 (1- cos 60)

          v = 29.4 m / s

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