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artcher [175]
3 years ago
14

When a flat slab of transparent material is placed under water, the critical angle for light traveling from the slab into water

is found to be 60°. What will the critical angle be if the slab is surrounded by air? Take the index of refraction for water to be 1.33. d) 45.6 c) 44.2 e) 47.3 a) 40.6 b) 42.5
Physics
1 answer:
Novosadov [1.4K]3 years ago
3 0

Answer:

(a) 40.6 degree

Explanation:

When refraction takes place from slab to water, the critical angle is 60 degree.

Use Snell's law

refractive index of water with respect to slab

\mu _{w}^{s}=\frac{Sin60}{Sin90}

\frac{\mu _{w}}{\mu _{s}}=0.866

\frac{1.33}{\mu _{s}}=0.866

μs = 1.536

Now for slab air interface, the critical angle is C.

\mu _{a}^{s}=\frac{SinC}{Sin90}

\frac{\mu _{a}}{\mu _{s}}=\frac{SinC}{Sin90}

1 / 1.536 = Sin C

C = 40.6 degree

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A parallel-plate capacitor is held at a potential difference of 250 V. A proton is fired toward a small hole in the negative pla
MissTica

Answer:

The speed of proton when it emerges through the hole in the positive plate is 2.05\times 10^5\ m/s.

Explanation:

Given that,

A parallel-plate capacitor is held at a potential difference of 250 V.

A A proton is fired toward a small hole in the negative plate with a speed of, u=3\times 10^5\ m/s

We need to find the speed when it emerges through the hole in the positive plate. It can be calculated using the conservation of energy as :

qV=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2\\\\1.6\times10^{-19}\times250=\dfrac{1}{2}mv^2-\frac{1}{2}\cdot1.67\times10^{-27}\cdot(3\times10^{5})^{2}\\\\\dfrac{1}{2}mv^2=3.515\cdot10^{-17}\\\\v=\sqrt{\dfrac{3.515\cdot10^{-17}\cdot2}{1.67\times10^{-27}}}\\\\v=2.05\times 10^5\ m/s

So, the speed of proton when it emerges through the hole in the positive plate is 2.05\times 10^5\ m/s.

5 0
3 years ago
What minimum speed must the block have at the base of the 70 m hill to pass over the pit at the far (right-hand) side of that hi
Drupady [299]

Answer:

initial velocity is v = 4.95 m / s

Explanation:

To solve this exercise we use the projectile launch ratios, when the block leaves the hill its speed is horizontal, let's find the time it takes to fall to the other point.

Initial vertical velocity is zero

          y = y₀ + v_{oy} t - ½ g t²

          y-y₀ = 0 -1/2 g t²

          t = \sqrt{ \frac{ 2(y_o -y)}{g} }

calculate

          t = \sqrt{ \frac{2 ( 70-50)}{9.8} }

          t = 2.02 s

with this time we can substitute in the horizontal displacement equation

          x = v₀ₓ t

          v₀ₓ = x / t

suppose that the distance between the two points is x = 10 m

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initial velocity is v = 4.95 m / s

4 0
3 years ago
The current world-record motorcycle jump is 77.0 m, set by Jason Renie. Assume that he left the take-off ramp at 13.0° to the ho
kotegsom [21]

Answer:

The take-off speed is 41.48 \frac{m}{s}

Explanation:

Given :

Range R = 77 m

Projectile angle \alpha  = 13°

From the formula of range,

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  v_{o} = 41.48 \frac{m}{s}

Therefore, the take-off speed is 41.48 \frac{m}{s}

7 0
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How many protons are in the radioactive isotope 40/19K?
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First we have to establish that the number of protons is equivalent to the atomic number of element. Here I am assuming that you are referring to Potassium (K) - 40. Potassium, stable or unstable has 19 protons.
8 0
3 years ago
I need help so bad pls pls help
erastovalidia [21]

Answer:

I think it is protection

6 0
2 years ago
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