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Natasha2012 [34]
4 years ago
7

Given the indices of refraction n_1 and n_2 of material 1 and material 2, respectively, rank these scenarios on the basis of the

phase shift in the refracted ray.
1) Rank from largest to smallest. To rank items as equivalent, overlap them.

a) n1,water=1.33
n2,air=1.00

b) n1,water=1.33
n2,quartz=1.46
c) n1,water=1.33
n2,diamond=2.42
d) n1,air=1.00
n2,quartz=1.46
e) n1,benzene=1.50
n2,water=1.33
f) n1,air=1.00
n2,water=1.33
2) Rank these scenarios on the basis of the phase shift in the reflected ray.
a) n1,benzene=1.50
n2,water=1.33
b) n1,air=1.00
n2,quartz=1.46
c) n1,water=1.33
n2,quartz=1.46
d) n1,water=1.33
n2,air=1.00
e) n1,water=1.33
n2,diamond=2.42
f) n1,air=1.00
n2,water=1.33
Physics
1 answer:
aniked [119]4 years ago
7 0

Answer:

a)  order of refraction is a, e, de, c , c) a f g

Explanation:

a) when lightning is refracted it must comply with the law of refraction

          .n₁ sin θ₁ = n₂ sinθ₂

          sin θ / sin δδa) when lightning is refracted it must comply with the law of refraction

          .n1 sin θ₁ = n2 sin θ ₂

          Sint θ1 / sin θ2 = n2 / n1

1 we see that the ray deviation is promotional to the ratio of the refractive indices

The order of refraction is a, e, de, c

.b) When a ray passes from a medium with less indicated refraction to a higher index the part of the reflected ray has a phase change of 180º

a) no phase change

b) reflected ray has a phase change of 180º

c) no phase change

d) no phase change

e) there is a phase change

f) these phase changeo  

1 we see that the ray deviation is promotional to the ratio of the refractive indices

The order of refraction is a, e, de, c

.b) When a ray passes from a medium with less indicated refraction to a higher index the part of the reflected ray has a phase change of 180º

a) no phase change

b) reflected ray has a phase change of 180º

c) no phase change

d) no phase change

e) there is a phase change

f) ay phase vabio

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Explanation:

Given that,

Force acting on the child, F = 310 N

Length of the ropes, d = 2.1 m

We need to find the gravitational potential energy of the child–Earth system relative to the child's lowest position when the ropes are horizontal. The potential energy is simply given by :

E=(mg)\times d\\\\E=F\times d\\\\E=310\ N\times 2.1\ m\\\\E=651\ J

Hence, this is the required solution.

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3 years ago
Object A of mass 0.70 kg travels horizontally on a frictionless surface at 20 m/s. It collides with object B, which is initially
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Answer:

25.71 kgm/s

Explanation:

Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.

Given that K₂ = 0.7K₁

1/2mv₂² = 0.7(1/2mv₁²)

v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s

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Pressure is about 1000 hPa at sea level and about 500 hPa at an altitude of 5.5 km. Why doesn’t this vertical pressure gradient
saul85 [17]

Answer:

A. The upward pressure gradient force is balanced by gravity.

Explanation:

A. is correct because the pressure difference is actually generated by gravity. As in the following formula for the pressure at different points:

p = p_0 + \rho g h

where p, p_0 are the pressure at 2 points, ρ is the density of the fluid, g is the gravitational constant, and h is the height difference.

B is incorrect because friction in air is too small to make an effect.

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6 0
3 years ago
Please help!!
pantera1 [17]

Answer:

(a) The time the ball stays in the air is approximately 4.66 seconds

(b) The distance from the cannon at which the ball will hit the ground is approximately 83.18 meters

(c) The maximum altitude of the ball is approximately 26.62 m

Explanation:

The given parameters of the question are;

The initial velocity at which a baseball cannon fires balls, u = 29 m/s

The angle at which the cannon is tilted, θ = 52°

(a) The time duration the ball stays in the air is given by the time of flight of the projected ball as follows;

2 \cdot t = \dfrac{2 \cdot u \cdot sin (\theta)}{g}

Where;

t = The time it takes the baseball to reach maximum height

2·t = The total time of flight = The time the ball stays in the air

θ = The angle at which the ball is tilted = 52°

g = The acceleration due to gravity ≈ 9.81 m/s²

u = The initial velocity = 29 m/s

Therefore, we have;

2 \cdot t = \dfrac{2 \times 29 \ m/s\times sin (52^{\circ})}{9.81 \ m/s^2} \approx 4.66 \ s

The time the ball stays in the air, 2·t ≈ 4.66 seconds

(b) The distance from the cannon at which the ball will hit the ground is given by the horizontal range, 'R', of the projectile as follows;

Horizontal \ range, R = \dfrac{u^2 \cdot sin(2 \cdot \theta) }{g}

∴ The distance from the cannon at which the ball will hit the ground = R

Horizontal \ range, R = \dfrac{(29 \ (m/s))^2 \cdot sin(2 \times 52^{\circ}) }{9.81 \ m/s^2} \approx 83.18 \, m

The distance from the cannon at which the ball will hit the ground = R ≈ 83.18 meters

(c) The maximum altitude of the ball is equal to the maximum height reached by the ball, H, which is given as follows;

H = \dfrac{u^2 \cdot sin^2 (\theta)}{2 \cdot g}

Therefore, we have;

H = \dfrac{(29 \ m/s)^2 \times sin^2 (52^{\circ})}{2 \times 9.81 \ m/s^2} \approx 26.62 \ m

The maximum altitude of the ball ≈ 26.62 m

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3 years ago
What could be used as another word for electrical potential
tekilochka [14]

Answer:

hey mate

answer is probably voltage as per me

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Explanation:

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