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ololo11 [35]
3 years ago
13

A beam of EMW in air strikes a sheet of a glass at 75 degrees with the normal in air. You observe that red light makes an angle

of 38.1 degrees with the normal in the glass, while violet light makes a 36.7 degree angle.
1. What are the indexes of refraction of this glass for these colors of light?
Answer in the order indicated. Separate your answers with a comma.
2. What are the speeds of red and violet light in the glass?
Physics
1 answer:
RUDIKE [14]3 years ago
5 0

Answer:

a) 1.57, 1.61

B 1.91x10^8m/s , 1.86 x 10^8m/s

Explanation:

Ok we know that refractive index is given as = sin i/ sin r

And i is the angle of incidence in the air and r is angle of refraction in the glass,

So we can sa y

for red light n = sin75/sin38.1 = 1.57

and for violet, n =sin75/sin36.7 = 1.61

So

(a) 1.57, 1.61

(b) we know that

Refractive index n = c/v,

where c is speed off light in air and v is speed of light in glass,

So , v = c/n

for red light v = c/1.57 = 1.91 x 10^8 m/s

and for violet light, v = c/1.61 =

1.86 x 10^8 m/s

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Moist air initially at 1258C, 4 bar, and 50% relative humidity is contained in a 2.5-m3 closed, rigid tank. The tank contents ar
brilliants [131]

Here is the missing part of the question

To Determine the heat transfer, in kJ  if the final temperature in the tank is 110 deg C

Answer:

Explanation:

The image attached below shows the process on T - v diagram

<u>At State 1:</u>

The first step is to find the vapor pressure

P_{v1} = \rho_1 P_g_1

= \phi_1 P_{x  \ at \ 125^0C}

= 0.5 × 232 kPa

= 116 kPa

The initial specific volume of the vapor is:

P_{v_1} v_{v_1} = \dfrac{\overline R}{M_v}T_1

116 \times 10^3 \times v_{v_1} = \dfrac{8314}{18} \times (125 + 273)

116 \times 10^3 \times v_{v_1} = 183831.7778

v_{v_1} = 1.584 \ m^3/kg

<u>At State 1:</u>

The next step is to determine the mass of water vapor pressure.

m_{v1} = \dfrac{V}{v_{v1}}

= \dfrac{2.5}{1.584}

= 1.578 kg

Using the ideal gas equation to estimate the mass of the dry air m_aP_{a1} V = m_a \dfrac{\overline R}{M_a}T_1

(P_1-P_{v1})  V = m_a \dfrac{\overline R}{M_a}T_1

(4-1.16) \times 10^5 \times 2.5 = m_a \dfrac{8314}{28.97}\times ( 125 + 273)

710000= m_a \times 114220.642

m_a = \dfrac{710000}{114220.642}

m_a = 6.216 \ kg

For the specific volume v_{v_1} = 1.584 \ m^3/kg , we get the identical value of saturation temperature

T_{sat} = 100 + (110 -100) \bigg(\dfrac{1.584-1.673}{1.210 - 1.673}\bigg)

T_{sat} =101.92 ^0\ C

Thus, at T_{sat} =101.92 ^0\ C, condensation needs to begin.

However, since the exit temperature tends to be higher than the saturation temperature, then there will be an absence of condensation during the process.

Heat can now be determined by using the formula

Q = ΔU + W

Recall that: For a rigid tank, W = 0

Q = ΔU + 0

Q = ΔU

Q = U₂ - U₁

Also, the mass will remain constant given that there will not be any condensation during the process from state 1 and state 2.

<u>At State 1;</u>

The internal energy is calculated as:

U_1 = (m_a u_a \ _{ at \ 125^0 C})+ ( m_{v1} u_v \ _{ at \ 125^0 C} )

At T_1 = 125° C, we obtain the specific internal energy of air

SO;

U_{a \ at \ 125 ^0C } = 278.93 + ( 286.16 -278.93) (\dfrac{398-390}{400-390}   )

=278.93 + ( 7.23) (\dfrac{8}{10}   )

= 284.714 \ kJ/kg\\

At T_1 = 125° C, we obtain the specific internal energy of  water vapor

U_{v1 \ at \ 125^0C} = u_g = 2534.5 \ kJ/kg

U_1 = (m_a u_a \ at \ _{  125 ^0C }) + ( m_{v1} u_v  \ at \ _{125^0C} )

= 6.216 × 284.714 + 1.578 × 2534.5

= 5768.716 kJ

<u>At State 2:</u>

The internal energy is calculated as:

U_2 = (m_a u_a \ _{ at \ 110^0 C})+ ( m_{v1} u_v \ _{ at \ 110^0 C} )

At temperature 110° C, we obtain the specific internal energy of air

SO;

U_{a \ at \ 110^0C } = 271.69+ ( 278.93-271.69) (\dfrac{383-380}{390-380}   )

271.69+ (7.24) (0.3)

= 273.862 \ kJ/kg\\

At temperature 110° C, we obtain the specific internal energy of  water vapor

U_{v1 \ at \ 110^0C}= 2517.9 \ kJ/kg

U_2 = (m_a u_a \ at \ _{  110 ^0C }) + ( m_{v1} u_v  \ at \ _{110^0C} )

= 6.216 × 273.862 + 1.578 × 2517.9

= 5675.57 kJ

Finally, the heat transfer during the process is

Q = U₂ - U₁

Q = (5675.57 - 5768.716 ) kJ

Q = -93.146 kJ

with the negative sign, this indicates that heat is lost from the system.

6 0
3 years ago
Question number 8 <br> Plz help
Dennis_Churaev [7]
To me, that sounds like the "Law of Conservation of Energy".
5 0
3 years ago
Use the diagram to explain how convection occurs inside the earth. What could convection cause to occur on earths surface
liubo4ka [24]

Answer:

mantle convection is the very slow creeping motion of earths solid silicate mantle caused by convection currents carrying heat from the interior to the planet's surface.

3 0
2 years ago
When a golf club hits a 0.0459 kg ball at rest, it exerts a 2380 N force for 0.00100 s. What is the speed of the ball afterwards
aksik [14]

Answer:

51.85m/s

Explanation:

Given parameters:

Mass of ball  = 0.0459kg

Force  = 2380N

Time taken  = 0.001s

Unknown:

Speed of the ball afterwards  = ?

Solution:

To solve this problem, we use Newton's second law of motion:

   F = m x \frac{v - u}{t}  

F is the force

m is the mass

v is the final velocity

u is the initial velocity

t is the time taken

        2380  = 0.0459 x \frac{v- 0}{0.001}  

        0.0459v  = 2.38

                   v = 51.85m/s

8 0
3 years ago
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a positively charged body makes contact with a body. after a while, the charged body becomes neutralised. state 3 main condition
MrRa [10]

Answer:

A body will become positively charged when some electrons will come out from the body.Thus, positive charge is due to deficiency of electrons.

5 0
2 years ago
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