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Mariulka [41]
3 years ago
9

As light shines from air to another medium, i = 26.0 º. The light bends toward the normal and refracts at 32.0 º. What is the in

dex of refraction? A. 1.06 B. 0.944 C. 0.827 D. 1.21
Physics
1 answer:
finlep [7]3 years ago
4 0

Explanation:

It is given that,

Angle of incidence from air to another medium, i = 26°

The angle of reflection, r = 32°

We need to find the refractive index of the medium. The ratio of sine of angle of incidence to the sine of angle of reflection is called refractive index. It can be given by :

n=\dfrac{\sin i}{\sin r}\\\\n=\dfrac{\sin (26)}{\sin (32)}\\\\n=0.82

So, the index of refraction is 0.82. Hence, the correct option is C.

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How do particles move when a transverse wave passes through a medium?
bagirrra123 [75]
<span> In </span>transverse waves<span>, </span>particles<span> of the</span>medium<span> vibrate </span>to<span> and from in a direction perpendicular </span>to<span> the direction of energy transport. </span>
4 0
3 years ago
7. 13 a turbine receives steam at 6 mpa, 600°c with an exit pressure of 600 kpa. Assume the turbine is adiabatic and neglect kin
AnnyKZ [126]

The work done by the turbine will be 708.2 kJ/kg. The work done by the turbine is the difference of the enthalpy at inlet and exit.

<h3 /><h3>What is temperature?</h3>

Temperature directs the hotness or coldness of a body. In clear terms, it is the method of finding the kinetic energy of particles within an entity. Faster the motion of particles, more the temperature.

If the given turbine is assumed to be reversible;

\rm P_I(Initial pressure)=60 mpa = 60 bar

\rm  T_i(Initial temperature)=600° C

\rm P_e (Exit pressure)=600 kpa=6 bar

The heat balance equation is;

\rm q-W_t=h_e-h_i\\\\ q=0\\\\\ W_t=h_i-h_e

The change in the entropy is;

\rm S_2-S_1=\frac{\delta q}{dt}

The work done by the turbine is;

\rm W_t = h_i-h_e\\\\ W_t =3658.4 -29.5019\\\\  W_t =708.2 \ kJ/kg

Hence,the work done by the turbine will be 708.2 kJ/kg.

To learn more about the temperature, refer to the link;

brainly.com/question/7510619

#SPJ4

6 0
2 years ago
If the charge on the negative plate of the capacitor is 121 nano-Coulomb, how many excess electrons are on that plate? Write you
Julli [10]

Answer:

n = 756.25 giga electrons

Explanation:

It is given that,

If the charge on the negative plate of the capacitor, Q=121\ nC=121\times 10^{-9}\ C

Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :

Q=ne

e is the charge on electron

n=\dfrac{Q}{e}

n=\dfrac{121\times 10^{-9}}{1.6\times 10^{-19}}

n=7.5625\times 10^{11}

or

n = 756.25 giga electrons

So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.

6 0
3 years ago
Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.
Leni [432]

Answer:

T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}

Explanation:

The given data :-

  • Diameter of rod AB ( d₁ ) = 200 mm.
  • Diameter of rod BC ( d₂ ) = 150 mm.
  • The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C
  • The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
  • Consider the required temperature increase to close the gap at C = T °C
  • Consider the change in length of the rod = бL

Solution :-

\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}

\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}

5 0
3 years ago
The earplug can reduce the sound level to about 18 decibels (dB). What percentage reduction is this intensity?
zlopas [31]

Answer:

1 x 10 -10 whisper at 1m distance.

Explanation:

  • Properly fitted ear plugs an reduce noise form 15-30db. Although they are better for low frequency
8 0
3 years ago
Read 2 more answers
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