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4 years ago

Under what circumstances will light display total internal reflection?

2 answers:

3 0

When light is passing from a denser medium to a lighter medium ( for example: from water to air)
<span> When the angle of refraction is greater than the critical angle( angle of incidence when the angle of refraction is perpendicular to the normal) of the denser medium.</span>

Lapatulllka [165]4 years ago

3 0

Answer:

light passing from a low-velocity medium to a high-velocity medium at an incident angle larger than the critical angle will be totally reflected.

Explanation:

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A 3.4kg aluminium ball has an apparent mass of 2.10kg when submerged in a particular liquid. Calculate the density of the liquid

Shalnov [3]

Answer:

1083.3kg/m ³

Explanation:

Given parameters:

Mass of aluminium ball = 3.4kg

Apparent mass of ball = 2.1kg

Unknown:

Density of the liquid = ?

Solution:

Density is is the mass per unit volume of any give substance;

Density = $\frac{mass}{volume}$

Now, we must understand that the apparent weight of the aluminium is the part of its weight supported by the fluid;

Pl = $\frac{Ml}{Vl}$

Pl = density of liquid

Ml = mass of liquid

Vl = volume of liquid

Mass of liquid = Mal - Mapparent

Mal = mass of aluminium

The volume of liquid displaced is the same as the volume of the aluminium according to Archimedes's principle;

Ml = Pl x Vl

Val = Vl

Val volume of aluminium

Vl = volume of liquid

****

Pal = $\frac{Mal}{Val}$

Val = $\frac{Mal}{Pal}$

Val = volume of aluminium

Mal = mass of aluminium

Pal = density of aluminium

*****

Since the Val = Vl

Ml = Pl x $\frac{Mal}{Pal}$

Since

Ml = Mal - Mapparent

Mal - Mapparent = Pl x $\frac{Mal}{Pal}$

Pal = density of aluminium = 2712 kg/m ³

3.4 - 2.1 = Pl x $\frac{3.4}{2712}$

1.3 = Pl x 0.00123

Pl = 1083.3kg/m ³

5 0

3 years ago

You push on the top edge of a 1.8 m tall solid circular cylinder of iron which is 4.00 cm in diameter. You push with a horizonta

Citrus2011 [14]

Answer:

$\triangle x=3.2*10^-^5 m$

Explanation:

From the question we are told that

Height of circular cylinder is $h= 1.8m$

Diameter of cylinder $D=4cm=>0.04m$

Horizontal Force $F=900N$

$Y = 10.0 *10^1^0 N/m^2 \\B = 9.0 * 10^1^0 N/m^2\\S = 4.0 * 10^1^0 N/m^2\\$

Generally the formula for shear modulus is mathematically represented by

$G=\frac{\tau}{\gamma}$

Where

$G=Shear modulus\\\tau= shear\ stress\\\gamma=shear strain$

$G=\frac{F/A}{\triangle x/L}$

$G=\frac{900/\pi r^2}{\triangle x/1.8}$

$4.0*10^1^0=\frac{900/\pi r^2}{\triangle x/1.8}$

$4.0*10^1^0=\frac{900}{\pi r^2} *\frac{1.8}{\triangle x}$

${\triangle x} =\frac{1620}{\pi r^2* 4.0*10^1^0}$

$\triangle x=3.2*10^-^5 m$

5 0

3 years ago

True or false: objects fall toward earth at a rate of 9.8 m/s because of centripetal force.

Citrus2011 [14]

Sadly, no. The statement kind of has some appropriate words in it, but it's badly corrupted. Objects don't fall to Earth at a rate of 9.8 m/s, and the force that accelerates them downward is not a centripetal one.

4 0

3 years ago

A particle oscillates in simple harmonic motion, with amplitude A and period T. The particle starts from position x = A. What is

frozen [14]

Answer:

i guess it is D because,

y = A sin(wt)

w = $\frac{2\pi }{T}$ and t = 3T/ 2

now, Y = A sin ( $\frac{2\pi }{T} X \frac{3T}{2}$ )

so, Y = 0radian

Explanation:

8 0

3 years ago

A 31 kg block is initially at rest on a horizontal surface. A horizontal force of 83 N is required to set the block in motion. A

Hatshy [7]

Answer:

The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.

Explanation:

By Newton's Laws of Motion and definition of maximum friction force, we derive the following two formulas for the static and kinetic coefficients of friction:

$\mu_{s} = \frac{f_{s}}{m\cdot g}$ (1)

$\mu_{k} = \frac{f_{k}}{m\cdot g}$ (2)

Where:

$\mu_{s}$ - Static coefficient of friction, no unit.

$\mu_{k}$ - Kinetic coefficient of friction, no unit.

$f_{s}$ - Static friction force, in newtons.

$f_{k}$ - Kinetic friction force, in newtons.

$m$ - Mass, in kilograms.

$g$ - Gravitational constant, in meters per square second.

If we know that $f_{s} = 83\,N$ , $f_{k} = 55\,N$ , $m = 31\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , then the coefficients of friction are, respectively: