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Marizza181 [45]
3 years ago
6

What is Snells Law? ...?

Physics
1 answer:
butalik [34]3 years ago
3 0
Snell's law<span> (also known as </span>Snell<span>–Descartes </span>law<span> and the </span>law<span> of refraction) is a formula used to describe the relationship between the angles of incidence and refraction, when referring to light or other waves passing through a boundary between two different isotropic media, such as water, glass, or air.


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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2 years ago
Can anyone help me with these questions? TIA!<br> (Don’t actually answer please! :) )
nataly862011 [7]
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Question 7: </h2>

 \huge\text{Graphs:}

The graph of  

• The I-V for Ohmic Metal wire conductor at constant temperature  always shows a straight line between the Current(I) plotted at Y axis and Voltage(V) plotted at X axis. Picture 1

• The I-V graph for Diode shows that first the current is zero but as we increase the potential difference(voltage), it results in the increase in the current. Picture 2

<h2>_____________________________________ </h2><h2>Question 8: </h2>

\Large\textbf{Diode:}  

A diode is a device that allows current to flow in only one direction.

\Large\textbf{Forward and Reverse Biasing:}  

Forward Bias, When a diode is forward bias (a voltage in the "forward" direction) then the P-side of the diode is attached to the positive terminal and N-side is fixed to the negative side of the battery which is connected, current flows freely through the device. The forward bias decreases the thickness of potential barrier(The potential barrier barrier in which the charge requires additional force for crossing the region)

Reverse Bias, When a diode is Reverse bias(a voltage in the "backward direction) then the P-side of the diode is connected to the negative terminal and N-side is connected to the positive terminal of the battery which is connected. The reverse bias increases the thickness of the potential barrier resulting in the flow of no current.

 \Large\textbf{Answer to the Question "Resistance"}

The Forward bias decreases the resistance of the diode whereas the reversed bias increases the resistance of the diode. As in forward biasing the current is easily flowing through the circuit whereas reverse bias does not allow the current to flow through it.

<h2>_____________________________________ </h2><h2>Best Regards, </h2><h2>'Borz' </h2>

8 0
3 years ago
Under which condition does Ohm's law apply? a. The current must be constant b. The power must be constant c. The temperature mus
ruslelena [56]

Answer:

option c

Explanation:

The temperature must be constant. Ohms law states that the current running through a conductor is directly proportional to the potential difference across it provided the temperature remains constant

6 0
2 years ago
A plane leaves Hartsfield Airport and flies around the world returning to Hartsfield airport What is the planes displacement?
rewona [7]
The way around the world
5 0
3 years ago
Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ
Angelina_Jolie [31]

Answer:

a). \frac{\dot{W}}{m}= 311 kJ/kg

b). \frac{\dot{\sigma _{gen}}}{m}=0.9113 kJ/kg-K

Explanation:

a). The energy rate balance equation in the control volume is given by

\dot{Q} - \dot{W}+m(h_{1}-h_{2})=0

\frac{\dot{Q}}{m} = \frac{\dot{W}}{m}+m(h_{1}-h_{2})

\frac{\dot{W}}{m}= \frac{\dot{Q}}{m}+c_{p}(T_{1}-T_{2})

\frac{\dot{W}}{m}= -30+1.1(980-670)

\frac{\dot{W}}{m}= 311 kJ/kg

b). Entropy produced from the entropy balance equation in a control volume is given by

\frac{\dot{Q}}{T_{boundary}}+\dot{m}(s_{1}-s_{2})+\dot{\sigma _{gen}}=0

\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+(s_{2}-s_{1})

\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+c_{p}ln\frac{T_{2}}{T_{1}}-R.ln\frac{p_{2}}{p_{1}}

\frac{\dot{\sigma _{gen}}}{m}=\frac{-30}{315}+1.1ln\frac{670}{980}-0.287.ln\frac{100}{400}

\frac{\dot{\sigma _{gen}}}{m}=0.0952+0.4183+0.3978

\frac{\dot{\sigma _{gen}}}{m}=0.9113 kJ/kg-K

5 0
3 years ago
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