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swat32
3 years ago
12

You observe a light ray move from one piece of glass to another (a different type of glass) and the light ray bends towards the

glass-glass interface (away from the normal) when it enters the 2nd piece of glass. You deduce that:
a. Glass 1 has a higher index of refraction than glass 2, and light moves faster in glass 2 than in glass 1
b. Glass 1 has a higher index of refraction than glass 2, and light moves faster in glass 1 than in glass 2
c. Glass 2 has a higher index of refraction than glass 1, and light moves faster in glass 1 than in glass 2
d. Glass 2 has a higher index of refraction than glass 1, and light moves faster in glass 2 than in glass 1
Physics
1 answer:
kodGreya [7K]3 years ago
3 0

Answer: C

Explanation: When light rays moves from one medium to another with a change in its direction (bending towards media interface), it is called refraction.

The angle the ray in the second medium (refracted ray) makes with the medium interface (normal) explains the bending of light and is dependent on the following the refractive index (n), wave speed in the medium (v) and other properties such as wavelength, and angle of incidence.

This question is focused on the relationship between refractive index and wave speed.

Refractive index (n) is inversely proportional to wave speed (v). This implies that a ray of light moving from a dense medium (say air) to a more dense medium (say glass) has it wave speed decreased and if reversed ( from glass to air) the wave speed increases.

A change in refractive index also affects the bending of the refracted ray.

A move from a dense to a more dense medium makes the refracted ray move towards the normal thus decreasing the angle of refraction (angle the refracted ray makes with the normal)

So for our question, since light ray (refracted ray) moves towards the glass to glass interface (normal) it means that light ray had moved from a dense to a more dense medium (that is glass 2 has higher refractive index than glass one) and the wave speed will decrease since there is an increase in refractive index (that is light travels faster in glass 1 than glass 2)

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Two stunt drivers drive directly toward each other. At time t=0 the two cars are a distance D apart, car 1 is at rest, and car 2
lesantik [10]

Answer: Hello there!

We know this:

The distance between the cars at t= 0 is D.

car 2 has an initial velocity of v0 and no acceleration.

car 1 has no initial velocity and a acceleration of ax that starts at  t = 0

then we could obtain the acceleration of the car 1 by integrating the acceleration over the time; this is v(t) = ax*t where there is not a constant of integration because the car 1 has no initial velocity.

Because the cars are moving against each other, we want to se at what time t they meet, this is equivalent to see:  

position of car 1 + position of car 2 = D

and in this way we could ignore constants of integration :D

for the position of each car we integrate again:  

P1(t) = (1/2)ax*t^2 and P2(t) = v0t

v0t + (1/2)ax*t^2 = D

v0t + (1/2)ax*t^2  - D = 0

now we can solve it for t using the Bhaskara equation.

t = \frac{-v0 +\sqrt{v0^{2} + 4*(1/2)ax*D } }{2(1/2)ax} =\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax}

that we cant solve witout knowing the values for v0, D and ax. But you could replace them in that equation and obtain the time, where you must remember that you need to choose the positive solution (because this quadratic equation has two solutions).

Now we want to know the velocity of car 1 just before the impact, this can be calculated by valuating the time in the as the time that we just found in the velocity equation for the car 1, this is:

v(\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax}) = ax*\frac{-v0 +\sqrt{v0^{2} + 2ax*D } }{ax} = {-v0 +\sqrt{v0^{2} + 2ax*D }

where again, you need to replace the values of v0, D and ax.

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Explain the difference between Longitudinal and Transverse waves.
masya89 [10]
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Suppose a wheel with a tire mounted on it is rotating at the constant rate of 3.37 times a second. A tack is stuck in the tire a
zhannawk [14.2K]

Answer:

The tangential speed of the tack is 8.19 m/s.

Explanation:

The wheel rotates 3.37 times a second that means wheel complete 3.37 revolutions in a second. Therefore, the angular speed ω of the wheel is given as follows:

\omega =3.37rev/s \times(\frac{2\pi rad}{1s} )\\\\=21.174rad/s

Use the relation of angular speed with tangential speed to find the tangential speed of the tack.

The tangential speed v of the tack is given by following expression

v = ω r

Here, r is the distance to the tack from axis of rotation.

Substitute 21.174 rad/s for ω, and 0.387 m for r in the above equation to solve for v.

v = 21.174 × 0.387

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4 0
3 years ago
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Which of the following is the flow of electrons through a wire or a conductor
Hatshy [7]
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8 0
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a sphere of diameter 6•0cm is moulded into a thin uniform wire of diameter 0•2mm.calculate the length of the wire in metres​
Scilla [17]

The length of the wire is 36 m.

<u>Explanation:</u>

Given, Diameter of sphere = 6 cm

We know that, radius can be found by taking the half in the diameter value. So,

       \text { sphere radius, } R=\frac{D}{2}=\frac{6}{2}=3 \mathrm{cm}=3 \times 10^{-2} \mathrm{m}

Similarly,

      \text { wire radius, } r=\frac{0.2}{2}=0.1 \mathrm{mm}=1 \times 10^{-3} \mathrm{m}

We know the below formulas,

          \text {volume of sphere}=\frac{4}{3} \times \pi \times R^{3}

          \text {volume of wire}=\pi \times r^{2} \times l

When equating both the equations, we can find length of wire as below, where \pi=\frac{22}{7}

          \frac{4}{3} \times \pi \times R^{3}=\pi \times r^{2} \times l

         \frac{4}{3} \times \frac{22}{7} \times\left(3 \times 10^{-2}\right)^{3}=\frac{22}{7} \times\left(1 \times 10^{-3}\right)^{2} \times l

The \pi value gets cancelled as common on both sides, we get

           \frac{4}{3} \times 27 \times 10^{-6}=10^{-6} \times l

The 10^{-6} value gets cancelled as common on both sides, we get

           l=4 \times 9=36 m

7 0
3 years ago
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