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kramer
3 years ago
8

Light incident on a lake surface is partly reflected and partly refracted.What is the differences between the reflected ray and

incident ray?
Physics
2 answers:
iren2701 [21]3 years ago
5 0

Answer: As per the question, a ray of light is incident on a surface and it is partly reflected and refracted. The incident light is an unpolarised light. The reflected light is partially polarised.

If the angle of incidence becomes equal to the Brester angle (polarising angle), then the reflected light is completely plane polarised.

nika2105 [10]3 years ago
5 0
The incident ray is a ray that hits the surface of the water. A reflected ray will always correspond to the incident ray and it is the light that is reflected by the surface of the water. Meaning, all rays that hit are incident rays, however, some are reflected rays.
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Which of the following would be the S I unit to use in measuring the temperature of a hot liquid a. Ampere b. Mole c. Celsius d.
rusak2 [61]

It would be C Celsius just took k12 test

5 0
3 years ago
Read 2 more answers
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
4. How long will it take a car travelling with a speed of 160 km hr to cover a distance of 700 meters? Hint: km/hr should be con
Inessa [10]

Answer:

15.8 seconds

Explanation:

Create an extended calculation to convert all the unit to what you need.

160 km      1000 m       1 hour         1 min

----------- x ------------- x -------------- x ----------   =  44.4 m/s

1 hour            1 km         60 min      60 sec

So 160km/hr is equal to 44.4m/s

Now you can figure out how many seconds it will take to go 700 meters.

44.4 m          

----------   X     x sec   =  700 m

1  sec

Solve for x sec

x sec = 700m / 44.4 m/s

         =  15.8 seconds

3 0
3 years ago
Which method of testing substances is a sure way to identify a chemical reaction?
Sergeu [11.5K]

Answer:

B

Explanation:

That's the answer. Hope it helped!

7 0
2 years ago
Please help me,its urgent!!
GREYUIT [131]

Test:

Performing a Litmus Test

Result:

Litmus paper gives the user a general indication of acidity or alkalinity as it correlates to the shade of red or blue that the paper turns.

  • To test the pH of a substance, dip a strip of litmus paper into the solution or use a dropper or pipette to drip a small amount of solution onto the litmus paper.
  • Blue litmus paper can indicate an acid with a pH between 4 and 5 or lower.
  • Red litmus paper can show a base with a pH greater than 8.
  • If a solution has a pH between 5 and 8, it will show little color change on the litmus paper.
  • A base tested with blue litmus paper will not show any color change, nor will an acid tested with red litmus paper register a change in color.
4 0
3 years ago
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