1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
patriot [66]
4 years ago
15

The fact that you can see interference colors in an oil slick on a puddle gives evidence that the oll film A) has two reflecting

surfaces B) is thin C) causes only 1 phase shift D) all of these E) two of these
Physics
1 answer:
luda_lava [24]4 years ago
7 0

Answer:

Option A

Explanation:

The interference color in an oil slick on a puddle results from the interference  caused as a result of reflection of light waves from the top as well as from the bottom surface of the oil slick.

This causes a disparity in the path of light reflecting from the surfaces and hence the sunlight is a band of seven colors, i.e., VIBGYOR, the disparities in the path results in constructive interference of the light rays which enables us to see interference colors in an oil slick.

You might be interested in
A train travels at 270 km in 3 hours . what is the speed of the train?
Ann [662]

270 \div 3 = 90km / h
3 0
3 years ago
Read 2 more answers
What is accerlation ?​
andre [41]

Explanation:

ability to gain speed over a short period of time.

the formula is:

_           v - v0               <Delta> v

a    =     _______    =   ________

                    t               <Delta> t

4 0
3 years ago
Read 2 more answers
Why is copper NOT used as a filament?
chubhunter [2.5K]

Answer:

Short answer, because copper wire does not have high resistance.

Explanation:

6 0
3 years ago
What is the relationship between thermal energy and temperature?
poizon [28]
As the air becomes warmer, heat<span> is transferred </span>between<span> molecules and kinetic</span>energy<span> is created which produces </span>thermal energy<span>. As the molecules move faster to transfer </span>heat<span>, the </span>temperature<span> also increases.</span>
6 0
3 years ago
A parallel plate capacitor is connected to a DC battery supplying a constant DC voltage V0= 600V via a resistor R=1845MΩ. The ba
tensa zangetsu [6.8K]

Answer:

See explanation

Explanation:

Given:-

- The DC power supply, Vo = 600 V

- The resistor, R = 1845 MΩ

- The plate area, A = 58.3 cm^2

- Left plate , ground, V = 0

- The right plate, positive potential.

- The distance between the two plates, D = 0.3 m

- The mass of the charge, m = 0.4 g

- The charge, q = 3*10^-5 C

- The point C = ( 0.25 , 12 )

- The point A = ( 0.05 , 12 )

Find:-

What is the speed, v, of that charge when it reaches point A(0.05,12)?

How long would it take the charge to reach point A?

Solution:-

- The Electric field strength ( E ) between the capacitor plates, can be evaluated by the potential difference ( Vo ) of the Dc power supply.

                           E = Vo / D

                           E = 600 / 0.3

                           E = 2,000 V / m

- The electrostatic force (Fe) experienced by the charge placed at point C, can be evaluated:

                           Fe = E*q

                           Fe = (2,000 V / m) * ( 3*10^-5 C)

                           Fe = 0.06 N

- Assuming the gravitational forces ( Weight of the particle ) to be insignificant. The motion of the particle is only in "x" direction under the influence of Electric force (Fe). Apply Newton's equation of motion:

                          Fnet = m*a

Where, a : The acceleration of the object/particle.

- The only unbalanced force acting on the particle is (Fe):

                          Fe = m*a

                          a = Fe / m

                          a = 0.06 / 0.0004

                          a = 150 m/s^2

- The particle has a constant acceleration ( a = 150 m/s^2 ). Now the distance between (s) between two points is:

                         s = C - A

                         s = ( 0.25 , 12 ) - ( 0.05 , 12 )

                         s = 0.2 m

- The particle was placed at point C; hence, velocity vi = 0 m/s. Then the velocity at point A would be vf. The particle accelerates under the influence of electric field. Using third equation of motion, evaluate (vf) at point A:

                        vf^2 = vi^2 + 2*a*s

                        vf^2 = 0 + 2*0.2*150

                        vf = √60

                        vf = 7.746 m/s

- Now, use the first equation of motion to determine the time taken (t) by particle to reach point A:

                       vf - vi = a*t

                       t = ( 7.746 - 0 ) / 150

                       t = 0.0516 s

- The charge placed at point C, the Dc power supply is connected across the capacitor plates. The capacitor starts to charge at a certain rate with respect to time (t). The charge (Q) at time t is given by:

                      Q = c*Vo*[ 1 - e^(^-^t^/^R^C^)]

- Where, The constant c : The capacitance of the capacitor.

- The Electric field strength (E) across the plates; hence, the electrostatic force ( Fe ) is also a function of time:

                     E = \frac{Vo*[ 1 - e^(^-^t^/^R^C^)]}{D} \\\\Fe = \frac{Vo*[ 1 - e^(^-^t^/^R^C^)]}{D}*q\\\\

- Again, apply the Newton's second law of motion and determine the acceleration (a):

                     Fe = m*a

                     a = Fe / m

                     a = \frac{Vo*q*[ 1 - e^(^-^t^/^R^C^)]}{m*D}

- Where the acceleration is rate of change of velocity "dv/dt":

                     \frac{dv}{dt}  = \frac{Vo*q}{m*D}  - \frac{Vo*q*[ e^(^-^t^/^R^C^)]}{m*D}\\\\B =  \frac{600*3*10^-^5}{0.0004*0.3} = 150, \\\\\frac{dv}{dt}  = 150*( 1 - [ e^(^-^t^/^R^C^)])\\\\

- Where the capacitance (c) for a parallel plate capacitor can be determined from the following equation:

                      c = \frac{A*eo}{d}

Where, eo = 8.854 * 10^-12  .... permittivity of free space.

                     K = \frac{1}{RC}  = \frac{D}{R*A*eo} =  \frac{0.3}{1845*58.3*8.854*10^-^1^2*1000} = 315\\\\

- The differential equation turns out ot be:

                     \frac{dv}{dt}  = 150*( 1 - [ e^(^-^K^t^)]) = 150*( 1 - [ e^(^-^3^1^5^t^)]) \\\\

- Separate the variables the integrate over the interval :

                    t : ( 0 , t )

                    v : ( 0 , vf )

Therefore,

                   \int\limits^v_0 {dv} \,  = \int\limits^t_0 {150*( 1 - [ e^(^-^3^1^5^t^)])} .dt \\\\\\vf  = 150*( t + \frac{e^(^-^3^1^5^t^)}{315} )^t_0\\\\vf = 150*( t + \frac{e^(^-^3^1^5^t^) - 1}{315}  )

- The final velocity at point A for the particle is given by the expression derived above. So for t = 0.0516 s, The final velocity would be:

                    vf = 150*( 0.0516 + \frac{e^(^-^3^1^5^*^0^.^0^5^1^6^) - 1}{315}  )\\\\vf = 7.264 m/s

- The final velocity of particle while charging the capacitor would be:

                   vf = 7.264 m/s ... slightly less for the fully charged capacitor

                     

7 0
3 years ago
Other questions:
  • When a ball is thrown straight up, by how much does the speed decrease each second? Neglect air resistance.
    13·1 answer
  • The mass number of an atom of carbon (C) represents the total number of
    8·1 answer
  • The teachers of Scott, Christopher, Dianne, and Kailee have last names Koeninger, Dannemiller, Briscoe, and Carter. Match the ki
    6·1 answer
  • The nervous system has two distinct branches. They are the:
    11·2 answers
  • A rocket, which is in deep space and initially at rest relative to an inertial reference frame, has a mass of 50.2 × 105 kg, of
    9·1 answer
  • Why do we need food even we lying in bed​
    10·2 answers
  • two wave pulses pass each other on a string. the one traveling toward the right has a positive amplitude, while the one travelin
    7·1 answer
  • Charge q1 is distance s from the negative plate of a parallel-plate capacitor. Charge q2=q1/3 is distance 2s from the negative p
    13·1 answer
  • Which is an example of how the body maintains homostasis
    11·2 answers
  • A circuit with a 60 V battery, a 7 Ω resistor, a resistor with 0.2 A across it, and another 7 Ω resistor in series. What is the
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!