Green light (525 nm) passes through a

Are there conditions under which the incident light ray undergoes reflection but not transmission at the boundary? if so, then w

Luden [163]

Total internal reflection causes light to be completely reflected across the boundary between the two media but not transmitted.

<h3>What is total internal reflection?</h3>

The term total internal reflection occurs when light is moving from a denser to a less dense medium such as from glass to air. This phenomenon occurs at the interface between the two media.

There are two conditions necessary for total internal reflection and they are;

1) Light must travel from a denser to a less dense medium

2) The angle of incidence in the denser medium must be greater than the critical angle.

Total internal reflection causes light to be completely reflected across the boundary between the two media but not transmitted.

Learn more about total internal reflection:brainly.com/question/13088998

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6 0

2 years ago

I WILL MARK BRAINLIEST IF CORRECT!!!

julsineya [31]

0.05*0.2=0.05*0.15+0.015*m2',

m2'=1/6 m/s,

m2'=0.17 m/s

N I C E - D A Y!

6 0

3 years ago

The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a

grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

$\sin\theta=\dfrac{3}{5}$

The horizontal component is

$T_{x}=T\cos\theta$

The vertical component is

$T_{y}=T\sin\theta$

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

$\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'$

Put the value into the formula

$0=190\times2+300\times 4-T\sin\theta\times 4$

$T\sin\theta\times 4=380+1200$

$T=\dfrac{1580\times5}{3\times 4}$

$T=658.33\ N$

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

$F_{x}=T\cos\theta$

Put the value into the formula

$F_{x}=658.33\times\dfrac{4}{5}$

$F_{x}=526.66\ N$

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

$F_{y}=F_{b}+F_{w}-T\sin\theta$

Put the value into the formula

$F_{y}=190+300-658.33\times\dfrac{3}{5}$

$F_{y}=95.002\ N$

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0

3 years ago

4) The object that changes its position relative to a fixed point with time is called ..<br>

mariarad [96]

Hold on lemme get the link for you

5 0

3 years ago

If the mass of an object is 8 kg and its momentum is -80 kgm/s, what is its velocity?

Dimas [21]

An object's momentum is the product of its mass and its velocity:

p = mv

p is its momentum, m is its mass, and v is its velocity.

Given values:

p = -80kg×m/s

m = 8kg

Plug in these values and solve for v:

-80 = 8v

v = -10m/s

Choice D

4 0

3 years ago