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Korolek [52]
3 years ago
12

A laser beam enters a 12.0 cm thick glass window at an angle of 33.0° from the normal. The index of refraction of the glass is 1

.41. At what angle from the normal does the beam travel through the glass? How long does it take the beam to pass through the plate?
Physics
1 answer:
Alenkinab [10]3 years ago
6 0

Answer:

The time is 0.563 ns.

Explanation:

Given that,

Index of refraction of glass = 1.41

Distance = 12.0 cm

Angle = 33.0°

We need to calculate the refraction angle

Using Snell's law

n_{1}\sin\theta=n_{r}\sin\theta

put the value into the formula

1\times\sin33=1.41\sin\theta

\sin\theta=\dfrac{1\times\sin33}{1.41}

\theta=22.71^{\circ}

We need to calculate the velocity of beam in glass

Using formula of velocity

v=\dfrac{c}{n}

Put the value into the formula

v=\dfrac{3\times10^{8}}{1.41}

v=2.13\times10^{8}\ m/s

We need to calculate the time

Using formula of distance

v=\dfrac{d}{t}

t=\dfrac{12.0\times10^{-2}}{2.13\times10^{8}}

t=5.63\times10^{-10}\ sec

t=0.563\times10^{-9}\ sec

t=0.563\ ns

Hence, The time is 0.563 ns.

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3 years ago
A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr
KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s

Time the parachutist falls without friction is 3.19 seconds

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2

Now the initial velocity of the last half height will be the final velocity of the first half height.

v=u+at\\\Rightarrow v=at

Since the height are equal

\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0

t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2

Magnitude of acceleration is -43200 ft/s²

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