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MA_775_DIABLO [31]

4 years ago

14

A diamond is cut such that the angle between its top surface and its bottom surface is α. For α=45∘, find the largest possible v

alue of the incident angle θa such that the blue light is totally internally reflected off the bottom surface. Express your answer in degrees to four significant figures.

1 answer:

Reil [10]4 years ago

4 0

Answer:

i = 61 degree

Explanation:

Given,

$\alpha =45^{o}$

Now, by the snell's law

$sin\theta_c=\frac{1}{n}\\sin\theta_c=\frac{1}{2.450}\\\theta_c=24.09^o$

Now,

Sin i / sin r = n 2 / n 1

sin i / sin r (45 - 24.09) = 2.45 / 1

i = 60.97 degree

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Find the quantity of heat needed

krok68 [10]

Answer:

Approximately $3.99\times 10^{4}\; \rm J$ (assuming that the melting point of ice is $0\; \rm ^\circ C$ .)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

$\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}$

The energy required comes in three parts:

Energy required to raise the temperature of that $0.100\; \rm kg$ of ice from $(-10\; \rm ^\circ C)$ to $0\; \rm ^\circ C$ (the melting point of ice.)
Energy required to turn $0.100\; \rm kg$ of ice into water while temperature stayed constant.
Energy required to raise the temperature of that newly-formed $0.100\; \rm kg$ of water from $0\; \rm ^\circ C$ to $10\;\ rm ^\circ C$ .

The following equation gives the amount of energy $Q$ required to raise the temperature of a sample of mass $m$ and specific heat capacity $c$ by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ ,

where

$c$ is the specific heat capacity of the material,
$m$ is the mass of the sample, and
$\Delta T$ is the change in the temperature of this sample.

For the first part of energy input, $c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}$ .

Similarly, for the third part of energy input, $c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}$ .

The second part of energy input requires a different equation. The energy $Q$ required to melt a sample of mass $m$ and latent heat of fusion $L_\text{f}$ is:

$Q = m \cdot L_\text{f}$ .

Apply this equation to find the size of the second part of energy input:

$\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}$ .

Find the sum of these three parts of energy:

$\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}$ .

3 0

3 years ago

When a falling meteoroid is at a distance above the Earth's surface of 3.40 times the Earth's radius, what is its acceleration d

Alchen [17]

Answer:

g = 0.85 m $s^{-2}$

Explanation:

g = $\frac{GM}{h^{2} }$

were; g is the acceleration due to Earth's gravity, G is Newton's gravitation constant (6.674 x $10^{-11}$ N $m^{2}$ $kg^{-2}$ ), M is the mass of the earth (5.972 x $10^{24}$ kg), and h is the distance of meteoroid to the earth.

h = 3.40 x R

= 3.40 x 6371 km

h = 21661.4 km

= 21661400 m

Thus,

g = $\frac{6.674*10^{-11}*5.972*10^{24} }{(21661400)^{2} }$

= $\frac{3.9857 *10^{14} }{4.6922*10^{14} }$

= 0.84944

g = 0.85 m $s^{-2}$

The acceleration due to the Earth's gravitation is 0.85 m $s^{-2}$ .

6 0

3 years ago

Compounds formed from the attraction of oppositely charged ions are called

Readme [11.4K]

Answer:

Ionic bond

Explanation:

Also called electrovalent bond, type of linkage formed from the electrostatic attraction between oppositely charged ions in a chemical compound.

Hope this helps! brainliest welcomed! :)

8 0

3 years ago

How to solve this step by step

-BARSIC- [3]

I think one hour, sorry if i'm wrong

8 0

3 years ago

Es muy urgente porfavor! ( solo me teneis que ayudar a sacar el apartado d,e y f porfavor!

Luba_88 [7]

I don’t know español

3 0

3 years ago