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FromTheMoon [43]

2 years ago

9

Light is incident normally on the short face of a 30∘−60∘−90∘ prism (Figure 1). A drop of liquid is placed on the hypotenuse of

the prism.
a) If the index of the prism is 1.50, find the maximum index that the liquid may have for the light to be totally reflected.
Express your answer using three significant figures.

1 answer:

ZanzabumX [31]2 years ago

7 0

1.06 is the <u>maximum</u> refractive index that the liquid may have for the light to be totally reflected.

Only when a light source passes from a denser to a rarer medium can it completely reflect.

When the angle of incidence surpasses a specific critical value, specular reflection occurs in the more highly refractive of the two mediums at their interface, and this reflection is known as total reflection.

sin $i_{c}$ = μ $_{r}$ / μ $_{d$

From the diagram

Angle of incidence = 60°

sin60° ≥ sin $i_{c}$ = μ $_{r}$ /μ $_{d}$

μ $_{r}$ ≤ μ $_{d}$ sin60°

μ $_{r}$ ≤ √1.5 × √3/2

= 1.06

Hence, the maximum index that the liquid may have for the light to be totally reflected is 1.06

Learn more about refractive index here brainly.com/question/10729741

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A 10.0-cm-long uniformly charged plastic rod is sealed inside a plastic bag. The net electric flux through the bag is 7.50 × 10

Rina8888 [55]

Answer:

66.375 x 10⁻⁶ C/m

Explanation:

Using Gauss's law which states that the net electric flux (∅) through a closed surface is the ratio of the enclosed charge (Q) to the permittivity (ε₀) of the medium. This can be represented as ;

∅ = Q / ε₀ -----------------(i)

Where;

∅ = 7.5 x 10⁵ Nm²/C

ε₀ = permittivity of free space (which is air, since it is enclosed in a bag) = 8.85 x 10⁻¹² Nm²/C²

Now, let's first get the charge (Q) by substituting the values above into equation (i) as follows;

7.5 x 10⁵ = Q / (8.85 x 10⁻¹²)

Solve for Q;

Q = 7.5 x 10⁵ x 8.85 x 10⁻¹²

Q = 66.375 x 10⁻⁷ C

Now, we can find the linear charge density (L) which is the ratio of the charge(Q) to the length (l) of the rod. i.e

L = Q / l ----------------------(ii)

Where;

Q = 66.375 x 10⁻⁷ C

l = length of the rod = 10.0cm = 0.1m

Substitute these values into equation (ii) as follows;

L = 66.375 x 10⁻⁷C / 0.1m

L = 66.375 x 10⁻⁶ C/m

Therefore, the linear charge density (charge per unit length) on the rod is 66.375 x 10⁻⁶ C/m.

3 0

2 years ago

9. Using these words explain how matter and energy interact when waves are generated. Disturbance ______________________________

ohaa [14]

Answer:

A wave is a disturbance of the space (or of a medium), that carries energy without transmitting matter.

A wave is produced when there is a periodic vibration in the particles of a medium (mechanical wave), or when there is a periodic oscillation of the electric and magnetic fields (electromagnetic waves). Electromagnetic waves are the only ones that can travel through a vacuum.

Mechanical waves are further classified into two types, depending on how the particles in the medium vibrate:

- If they vibrate up and down (perpendicular to the direction of motion of the wave), they are called transverse waves

- If they vibrate back and forth (parallel to the direction of motion of the wave), they are called longitudinal waves

In general, waves are generated from a precise point in the space, which is called source of the wave. The source of the wave does work, since it is responsible for starting the motion of the particle, and make them starting vibrating, so it transmits energy to the particles.

7 0

3 years ago

A 55.0-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300. What h

MariettaO [177]

Answer:

161.86 N

Explanation:

mass of box m= 55.0 kg

weight of the box, mg= 55×9.81

g here is acceleration due to gravity =9.81 m/sec^2

coefficient of friction between the box and the surface μ= 0.3

the friction force F_s= μmg= 0.3×55×9.81

=161.86 N

to move the ball horizontal force required is 161.86 N

8 0

2 years ago

A student on an amusement park ride moves in circular path with a radius of 3.5 m once every 4.5 s. What is the tangential veloc

nata0808 [166]

Answer:

the tangential velocity of the student is 4.89 m/s.

Explanation:

Given;

the radius of the circular path, r = 3.5 m

duration of the motion, t = 4.5 s

let the student's tangential velocity = v

The tangential velocity of the student is calculated as follows;

$v = \frac{2\pi r}{t} \\\\v = \frac{2 \pi \times 3.5}{4.5} \\\\v = 4.89 \ m/s$

Therefore, the tangential velocity of the student is 4.89 m/s.

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2 years ago

A transverse wave on a string is described by the following wave function.y = (0.090 m) sin (px/11 + 4pt)(a) Determine the trans

alukav5142 [94]

Explanation:

(a) It is known that equation for transverse wave is given as follows.

y = $(0.09 m)sin(\pi \frac{x}{11} + 4 \pi t)$

Now, we will compare above equation with the standard form of transeverse wave equation,

y = $A sin(kx + \omega t)$

where, A is the amplitude = 0.09 m

k is the wave vector = $\frac{\pi}{11}$

$\omega$ is the angular frequency = $4\pi$

x is displacement = 1.40 m

t is the time = 0.16 s

Now, we will differentiate the equation with respect to t as follows.

The speed of the wave will be:

v(t) = $\frac{dy}{dt}$

v(t) = $A \omega cos(kx + \omega t)$

v(t) = $(0.09 m)(4\pi) cos(\frac{\pi \times 1.4}{11} + 4 \pi \times 0.16)$

v(t) = -0.84 m/s

The acceleration of the particle in the location is

a(t) = $\frac{dv}{dt}$

a(t) = $-A \omega 2sin(kx + \omega t)$

a(t) = $-(0.09 m)(4 \pi)2 sin(\frac{\pi \times 1.4}{11} + 4\pi \times 0.16)$

a(t) = -9.49 $m/s^{2}$

Hence, the value of transverse wave is 0.84 m/s and the value of acceleration is 9.49 $m/s^{2}$ .

(b) Wavelength of the wave is given as follows.

$\lambda = \frac{2\pi}{k}$

$\lambda = (frac{2\pi}{\frac{\pi}{11})
$

$\lambda$ = 22 m

The period of the wave is

T = $\frac{2 \pi}{\omega}$

T = $\frac{2 \pi}{4 \pi}$

= 0.5 sec

Now, we will calculate the speed of propagation of wave as follows.

v = $\frac{\lambda}{T}$

= $\frac{22 m}{0.5 s}$

= 44 m/s

therefore, we can conclude that wavelength is 22 m, period is 0.5 sec, and speed of propagation of wave is 44 m/s.

7 0

3 years ago