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2 years ago

the speed of light in a piece of plastic is 2.00x10^8 meters per second what is the absolute index of refraction of this plastic

1 answer:

6 0

The speed of light in a material with refractive index n is modified according to the following law:
$v= \frac{c}{n}$
where c is the speed of light in vacuum ( $c=3\cdot 10^8 m/s$ ), while v is the speed of light in the medium ( $v=2 \cdot 10^8 m/s$ ). Re-arranging, we can find the refractive index of this plastic:
$n= \frac{c}{v}= \frac{3 \cdot 10^8 m/s}{2 \cdot 10^8 m/s}=1.5$

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A composite wall separates combustion gases at 2400°C from a liquid coolant at 100°C, with gas and liquid-side convection coeffi

evablogger [386]

Answer:

$\text{heat loss} = 24864.05 \ W/m^2$

Explanation:

$T_1$ , $T_2$ are temperatures of gasses and liquid in Kelvins,

$t_1$ and $t_2$ are thicknesses of gas layer and steel slab in meters,
$h_1$ , $h_2$ are convection coefficients gas and liquid in $W/m^2 \cdot K$ ,
$R_c$ is the contact resistance in $m^2 \cdot K/W$ ,

and $k_1, k_2$ are thermal conductivities of gas and steel in $W/m \cdot K$ ,

then: part(a):

$\text{heat loss } = \frac{T_1 - T_2} { \frac{1}{h_1} + \frac{t_1}{t_2} + R_c + \frac{t_2}{k_2} + \frac{1}{h_2}}$

using known values:

$\text {heat loss} = 2486.05 W/m^2$

part(b): Using the rate equation :

$\text {heat loss} = h_1 (T_1 - T_{s1})$

the surface temperature $T_{s1} = 1678.438 \ K$

and $T_{c1} = T_{s1} - \frac {t_1 (\text{heat loss})}{k_1} = 1664.560 \ K$

Similarly

$T_{c2} = T_{c1} - R_c (\text{heat loss}) = 421.357 \ K$

$T_{s2} = T_{c2} - \frac {t_2 (\text{heat loss})}{ k_2} = 397.864 \ K$

The temperature distribution is shown in the attached image

3 0

2 years ago

An electric buzzer is activated, then sealed inside a glass chamber. When all of the air is pumped out of the chamber, how is th

Aleksandr-060686 [28]

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2 years ago

flat block is pulled along a horizontal flat surface by a horizontal rope perpendicular to one of the sides. The block measures

Lelu [443]

Answer:

Answered

Explanation:

v= 1 m/s

A= 1 m^2

m= 100 kg

y= 1 mm

μ = ?

ζ= viscosity of SAE 20 crankcase oil of 15° C= 0.3075 N sec/m^2

forces acting on the block are

F_s ← ↓ →F_f

N= mg

F_s= shear force = ζAv/y F_f= friction force = μN

now in x- direction F_s= F_f

ζAv/y = μN

0.3075×1×1×1/1×10^{-3} = μ×100

⇒μ=0.313 (coefficient of sliding friction for the block)

Now, as the velocity is increased shear force also increases and due to this frictional force also increases.

Now, to compensate this frictional force friction coefficient must increase

as v∝μ

7 0

2 years ago

Read 2 more answers

Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of

schepotkina [342]

If speed = distance/time , then time = speed/distance.

So...

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Average distance from Earth to Sun = 149.6*10^9(m)

Therefore, t=(3*10^8(m/s))/(149.6*10^9(m))

I hope this was a helpful explanation, please reply if you have further questions about the problem.

Good luck!

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2 years ago

A real heat engine operates between temperatures tc and th. during a certain time, an amount qc of heat is released to the cold

tino4ka555 [31]

$q_{c}$ = Heat released to cold reservoir