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3 years ago

I want explanation with the correct answer as to why the answer is 0 degree despite refraction of light .

Physics

1 answer:

Alona [7]3 years ago

4 0

Answer:

Refraction only occurs when the light is incident to the surface at some positive angle. When light is perpendicular to a surface there is no bend to the ray.

Using Snell's law, the math looks like this

n₁sinθ₁ = n₂sinθ₂

at an incidence of 0°

n₁sin(0) = n₂sinθ₂

0 = n₂sinθ₂

as n₂ is always a positive number, sinθ₂ must equal 0 meaning θ₂ = 0

So while there will be refraction of light a position O, which is the exact center of the semicircular prism and the refracted ray will follow a radial line, there will be no refraction of light at P because the light is traveling perpendicular to the surface there. There will be a change in light velocity because of the change in the medium, but no refraction.

Explanation:

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Do atoms ever touch

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Answer:

nope never

Explanation:

8 0

3 years ago

Read 2 more answers

A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. how long does h

noname [10]

The shot putter should get out of the way before the ball returns to the launch position.

Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.

The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 = 0.45 s

t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.

Answer: 0.45 s

7 0

3 years ago

The bigclaw snapping shrimp shown in (Figure 1) is aptly named--it has one big claw that snaps shut with remarkable speed. The p

leva [86]

1) $1.86\cdot 10^6 rad/s^2$

2) 2418 rad/s

3) $27000 m/s^2$

4) 36.3 m/s

Explanation:

The angular acceleration of an object in rotation is the rate of change of angular velocity.

It can be calculated using the following suvat equation for angular motion:

$\theta=\omega_i t +\frac{1}{2}\alpha t^2$

where:

$\theta$ is the angular displacement

$\omega_i$ is the initial angular velocity

t is the time

$\alpha$ is the angular acceleration

In this problem we have:

$\theta=90^{\circ} = \frac{\pi}{2}rad$ is the angular displacement

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

Solving for $\alpha$ , we find:

$\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2$

For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

$\omega_f = \omega_i + \alpha t$

where

$\omega_i$ is the initial angular velocity

t is the time

$\alpha$ is the angular acceleration

In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

Therefore, the final angular speed is:

$\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s$

The tangential acceleration is related to the angular acceleration by the following formula:

$a_t = \alpha r$

where

$a_t$ is the tangential acceleration

$\alpha$ is the angular acceleration

r is the distance of the point from the centre of rotation

Here we want to find the tangential acceleration of the tip of the claw, so:

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

$a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2$

Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:

$v=u+at$

where

u is the initial linear speed

a is the tangential acceleration

t is the time elapsed

Here we have:

$a=27900 m/s^2$ (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

$v=0+(27900)(0.0013)=36.3 m/s$

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4 years ago

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4 years ago

If a road does not have a bicycle lane, where must a bicyclist ride their bicycle?

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A bicyclist can ride their bicycle still on the road. Bicycle riders be able to take the public ways which has the similar rights and accountability as motorists and are subject to the same guidelines and protocols. The law says that individuals who ride bikes should ride as nearby to the right side of the road as likely excluding under the following conditions: when passing, preparing for a left go, evading risks, if the lane is too constricted to share, or if oncoming a place where a right turn is approved. In a road which has a bike lane the bicyclists roving slower than road traffic must custom the bike way excluding when creating a left turn, passing, evading hazardous settings, or impending a place where a right turn is approved.

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4 years ago