All categories

3 years ago

What is Snell's Law?

Physics

1 answer:

polet [3.4K]3 years ago

3 0

Answer:

law stating that the ratio of the sines of the angles of incidence and refraction of a wave are constant when it passes between two given media.

Snell's law, in optics, a relationship between the path taken by a ray of light in crossing the boundary or surface of separation between two contacting substances and the refractive index of each. ... Snell's law asserts that n1/n2 = sin α2/sin α1.

You might be interested in

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago

Which illustrates projectile motion?

Step2247 [10]

Answer:

Ans is (B) driving a rock from a building

7 0

3 years ago

What’s is the movement of one object around another

allsm [11]

Answer:

revolution

Explanation:

8 0

4 years ago

a ball is thrown upward with initial velocity of 20 m/s. (a) how long is the ball in the air? (b) what is the greatest height re

Lena [83]

a) t = 4.08 s

b) Response = 20.39m

c) t = 0.991 and 3.087 seconds.

Initially, the vertical speed was 20 m/s.

(a) how long is the ball in the air?

The ball is tossed upward, rises to its highest point, and then hits the ground again.

Consequently, the end height will be equal to the starting height.

h = h o = Om

Use the equation s= ut + -at.

2764 – An = 14

Put a negative sign for g because the ball is being thrown upwards and g is acting downwards.

Om = t - 0.5 * 9.81 m/s * 20 m/s

0 = 20 *t - 0.5 * 9.81 *t

20 *t=0.5*9.81 *t

20 = 0.5 * 9.81 *t

20 0.5*9.81

Answer (a): t = 4.08 s

(b) what is the greatest height reached by the ball?

Vertical velocity for the ball decreases to zero when it reaches its highest point.

Use the equation 2-u=295

-2ghmas for - u

(20m/s - (Om/s)2) = -2 * 9.81 m/s2 *

02 – 20= -2*9.81 *

-20% = -2 *9.81 *

202 = 2 * 9.81 *

b) RESPONSE: H = 20.39m

(c) when is the ball 15 m above the ground?

The ball will be 15 meters above the earth on two separate occasions.

1. When climbing

2. When descending

Use the equation s= ut + -at.

2,264 – fn = 4

0.5 * 9.81 m/s * 15m = 20 m/s*t

15 = 20 *t - 0.5 * 9.81 *t

15 = 20t – 4.9057

4.905 – 20t + 15 = 0

Calculator-based quadratic equation solution

(c) t = 0.991 and 3.087 seconds.

Hence the answers are,

a) t = 4.08 s

b) RESPONSE: Hmar = 20.39m

c) t = 0.991 and 3.087 seconds.

To learn more about initial velocity, click brainly.com/question/14154244

#SPJ4

4 0

2 years ago

car travels 13 meters per second accelerates to 25 meters per second, how far did it travel in 6 seconds

Sergio039 [100]

It traveled 150 meters

7 0

3 years ago

What is Snell's Law?​

What is Snell's Law?