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Allisa [31]
3 years ago
10

. A laser beam shines along the surface of a block of transparent material (see Fig. E33.8). Half of the beam goes straight to a

detector, while the other half travels through the block and then hits the detector. The time delay between the arrival of the two light beams at the detector is 6.25 ns. What is the index of refraction of this material
Physics
1 answer:
Neporo4naja [7]3 years ago
6 0

Answer:

n = 1,875

Explanation:

The speed of light in vacuum is constant (c) and in a material medium it is

          v = d / t

The refractive index of a material is defined by

         n = c / v

         

Let's look for the speed of light in the material, in general the length that light travels is known, this value is high, x = 1, when we place a block on the road, a small amount is lengthened by the length of the block, which in general is despised

These measurements are made on a digital oscilloscope that allows to stop the signals and measure their differences, that is, the zero is taken when the first ray arrives and the time for the second ray is measured,

         

         v = d / t

         v = 1 / 6.25 10⁻⁹

         v = 1.6 10⁸ m / s

we calculate the refractive index

        n = 3 10⁸ / 1.6 10⁸

        n = 1,875

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After six half-lives, what percentage of a radioactive sample will remain?
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Suppose you have a 115 kg wooden crate resting on a wood floor, with coefficient of static friction 0.500 between these wood sur
alexdok [17]

Answer:

maximum force (in N) can you exert horizontally on the crate without moving it = 564.075 N

Explanation:

Given data:

mass m= 115 kg

coefficient of friction μ =0.5

according to newton law of motion the net force acted on body is given by

F_{net}=ma=F_{max}- F_{f}

here the body is in rest ,

F_{net}=0\\\Rightarrow F_{max}=F_{f}\\\text{frriction force}  F_{f} = \mu\times N\\

N= it is the normal reaction force

N= mg = 115×9.81

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Define what is false​
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3 years ago
Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th
DiKsa [7]

Answer:

a.\rm -1.49\ m/s^2.

b. \rm 50.49\ m.

Explanation:

<u>Given:</u>

  • Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

<h2>(a):</h2>

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).

At time t = 3 seconds,

\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

<h2>(b):</h2>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.

For the time interval of 2 seconds,

\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.

It is the displacement of the particle in 2 seconds.

7 0
3 years ago
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