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2 years ago

What are the properties of glass in relation to microscopy?

2 answers:

8 0

Answer: When a glass is broken, some of the resulting fragments will retain an original surface. Light microscopy can be used to identify these surfaces and to examine surface features characteristic of the type and origin of the glass. The techniques employed are specular reflection, fluorescence, Nomarski differential interference contrast microscopy and interferometry using a Hartley objective.

Alekssandra [29.7K]2 years ago

7 0

Answer:

<h3>Following are the properties and characteristics of the glass.</h3>

<h3>Hardness and Brittleness. It is a hard material as it has great impact resistance against applied load.</h3>

Weather Resistance.
Insulation.
Chemical Resistance.
Colour and Shape Varieties.
Transparency.
Fire Resistant Glazing.
Property Modification.

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The hydraulic oil in a car lift has a density of 8.53 x 102 kg/m3. The weight of the input piston is negligible. The radii of th

navik [9.2K]

Answer:

(a) the input force is 36.56 N

(b) the input force is 37.49 N

Explanation:

Given;

density of hydraulic oil, ρ = 8.53 x 10² kg/m³

radius of plunger, r₁ = 0.135 m

radius of piston, r₂ = 5.43 x 10⁻³ m

Part (a) The input force needed to support 22600-N weight, when the bottom surfaces of the piston and plunger are at the same level;

$P =\frac{F}{A}$

Where;

P is pressure

F is force

A is circular area = πr²

$\frac{F_1}{A_1} =\frac{F_2}{A_2} \\\\F_2 = \frac{F_1*A_2}{A_1} =\frac{F_1* \pi r_2^2}{\pi r_1^2} = \frac{F_1* r_2^2}{ r_1^2} \\\\F_2 = \frac{22600*(5.43*10^{-3})^2 }{(0.135)^2}\\\\F_2 = 36.56 \ N$

Part (b) The input force needed to support 22600-N weight, when the bottom surface of the output plunger is 1.20 m above that of the input plunger

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But, F = PA and A = πr²

$F_2 = F_1(\frac{A_2}{A_1} ) + \rho gh*A_2\\\\F_2 = F_1(\frac{r_2^2}{r_1^2} )+\rho gh(\pi r_2^2)\\\\F_2 = 22600(\frac{5.43*10^{-3}}{0.135})^2 \ + 853*9.8*1.2*\pi (5.43*10^{-3})^2\\\\F_2=36.56 + 0.93\\\\F_2 = 37.49 \ N$

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3 years ago