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uranmaximum [27]
2 years ago
15

Suppose you have an unknown clear substance immersed in water, and you wish to identify it by finding its index of refraction. Y

ou arrange to have a beam of light enter it at an angle of 45.7∘, and you observe the angle of refraction to be 30.5∘. What is the index of refraction of the substance? Water has an index of refraction equal to 1.333.
Physics
1 answer:
s344n2d4d5 [400]2 years ago
3 0

Answer: 1.65

Explanation: Snell's law says that index of refraction of the first material times sine theta one equals index of refraction of the second material, which we don't know, multiplied by sine theta two, . So, we'll solve this for N two by dividing both sides by sine theta two. So, N two is N one sine theta one over sine theta two . So, that's 1.33 index of refraction of water times sine of 45.7degrees incidence angle divided by sine of 30.5 degrees angle of refraction, giving us 1.65.

The formula for snell's law is N1*sintheta1=N2*sinetheta2

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Two parallel plates that are initially uncharged are separated by 1.7 mm, have only air between them, and each have surface area
yaroslaw [1]

Answer:

5.63\cdot 10^{-6} C

Explanation:

The capacitor of a parallel-plate capacitor is given by:

C=\epsilon_0 \frac{A}{d}

where

A is the area of each plate

d is the separation between the plates

\epsilon_0 is the vacuum permittivity

The energy stored in a capacitor instead is given by

U=\frac{1}{2}\frac{Q^2}{C}

where

Q is the charge stored in each plate

Substituting the expression we found for C inside the last formula,

U=\frac{1}{2}\frac{Q^2 d}{\epsilon_0 A}

And re-arranging it

Q=\sqrt{\frac{2U\epsilon_0 A}{d}}

Now if we substitute

d=1.7 mm=0.0017 m\\A=16 cm^2 = 16\cdot 10^{-4} m^2\\U = 1.9 J

We find the charge stored on the capacitor:

Q=\sqrt{\frac{2(1.9)(8.85\cdot 10^{-12})(16\cdot 10^{-4})}{0.0017}}=5.63\cdot 10^{-6} C

7 0
3 years ago
An object is dropped and falls freely to the ground with an acceleration of g. If it is thrown upward at an angle instead, its a
Savatey [412]

Answer:

g

Explanation:

if an object is thrown upward or at any angle, the acceleration acting on that object is the same as acceleration due to gravity which always acts towards the vertically downwards direction because there is no acceleration or the force acting on the object in horizontal direction.

Thus, the acceleration is same as acceleration due to gravity g.  

5 0
3 years ago
A localized impediment to electron flow in a circuit is a:
hichkok12 [17]
This would be the definition of a resistor. These components inhibit or “resist” the flow of a current.

Hope this helps!
8 0
3 years ago
Read 2 more answers
As sea floor spreading occurs, the oceanic plate _______.
Ira Lisetskai [31]

Becomes older

Explanation:

As sea floor spreading occurs at divergent margins, the oceanic plate becomes older. Younger plate margin are the closest to the margin whereas the older plates bushes backward away from the spreading centers.

  • The idea that the sea floor spreads was postulated by Harry Hess shortly after the second world war around the 1960's.
  • At divergent margins new crust materials from the mantle are brought to the surface.
  • They crystallize and settle at the flanks of plate margins.
  • Older ones are pushed backward away from the margin into far away subduction zones.

Learn more:

Sea floor spreading brainly.com/question/9912731

#learnwithBrainly

8 0
3 years ago
Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton
solmaris [256]

Answer:

a

\lambda = 3.68 *10^{-36} \  m

b

\lambda_p = 1.28*10^{-14} \ m

Explanation:

From the question we are told that

   The mass of the person is  m =  180 \  kg

    The speed of the person is  v  =  1 \  m/s

    The energy of the proton is  E_ p =  5 MeV = 5 *10^{6} eV  = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \  J

Generally the de Broglie wavelength is mathematically represented as

      \lambda = \frac{h}{m * v }

Here  h is the Planck constant with the value

      h = 6.62607015 * 10^{-34} J \cdot s

So  

     \lambda = \frac{6.62607015 * 10^{-34}}{ 180  * 1  }

=> \lambda = 3.68 *10^{-36} \  m

Generally the energy of the proton is mathematically represented as

         E_p =  \frac{1}{2}  *   m_p  *  v^2_p

Here m_p  is the mass of proton with value  m_p  =  1.67 *10^{-27} \  kg

=>     8.0*10^{-13} =  \frac{1}{2}  *   1.67 *10^{-27}  *  v^2

=>   v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }

=>   v = 3.09529 *10^{7} \  m/s

So

        \lambda_p = \frac{h}{m_p * v_p }

so    \lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }

=>     \lambda_p = 1.28*10^{-14} \ m

     

5 0
3 years ago
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