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Vilka [71]
3 years ago
9

A thin glass rod is submerged in oil. What is the critical angle for light traveling inside the rod? The index of refraction for

glass is 1.50 The index of refraction for oil is 1.46.
Physics
1 answer:
Alekssandra [29.7K]3 years ago
7 0
When light travels from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle of incidence above which light is reflected only (no refraction occurs), and the value of this critical angle is given by
\theta_c = \arcsin ( \frac{n_2}{n_1} )
where n2 is the refractive index of the second medium and n1 is the refractive index of the first medium.

In this problem, the first medium is the glass (n_1 = 1.50), while the second medium is oil (n_2 =1.46), therefore the critical angle is given by
\theta_c = \arcsin( \frac{1.46}{1.50} )=\arcsin(0.973)=76.7^{\circ}
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Take a look at a bicycle with gears. Using what you have just learned, answer the following questions:
Harlamova29_29 [7]

Answer:

answer a: a large front gear with a small back gear

answer b: a small front gear with a large back gear

Explanation:

just simple gearing ratios

4 0
2 years ago
A certain type of bird lives in two regions of a state. The distribution of weight for birds of this type in the northern region
Annette [7]

Answer: (a) Z-score are 1 and -1.2 for northern and southern regions, respectively.

Explanation: <u>Z-score</u> is how many standard deviations a data is from the population mean or how far a data point is from the mean.

The z-score is calculated by the following:

z=\frac{x-\mu}{\sigma}

where

x is the data point

μ is population mean

σ is standard deviation

For the <u>northern</u> <u>region</u> birds:

μ = 10, σ = 3, x = 13

z=\frac{13-10}{3}

z = 1

The z-score for birds living in the northern region is 1, which means it is 1 standard deviation <em>above the mean</em>.

For the southern region:

μ = 16, σ = 2.5, x = 13

z=\frac{13-16}{2.5}

z = -1.2

The z-score for southern living birds is -1.2, meaning it is 1.2 standard deviations <em>below the mean</em>.

6 0
3 years ago
The two cyclists travel at the same speed on level ground. They approach a low hill and decide to coast up instead of hard pedal
Sergeeva-Olga [200]

Answer:

  • <u>Option-(E): </u>Bike B; its wheels have a smaller moment of inertia.                                                                                                                                                                                                                                                                              

Explanation:

  • The two cyclists travel at the same speed on level ground, as they approach a low hill and decide to coast up instead of hard pedaling. At the top of the hill, as bike "B" will have a larger speed, as compared to the bike "A".Due, to the effect of having smaller moment of inertia, I=m×r².
  • Having no friction nor air resistance, and all the wheels roll on the ground without slipping. While, the B have larger speed,v due to the fact that bike "B" will have a larger speed,"v" as compared to the bike "A".
3 0
3 years ago
A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an ang
adell [148]

Answer:

A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an angle of 36.9 ∘ above the horizontal. The monitor's speed is a constant 2.30 cm/s.

how much work is done on the monitor by (a) friction, (b) gravity

work(friction) = 453.5J

work(gravity) = -453.5J

Explanation:

Given that,

mass = 14kg

displacement length = 5.50m

displacement angle = 36.9°

velocity = 2.30cm/s

F = ma

work(friction) = mgsinθ .displacement

                      = (14) (9.81) (5.5sin36.9°)

                       = 453.5J

work(gravity)

= the influence of gravity oppose the motion of the box and can be pushing down, on the box from and angle of (36.9° + 90°)

= 126.9°

work(gravity) = (14) (9.81) (5.5cos126.9°)

                      = -453.5J

8 0
2 years ago
A wheel with a tire mounted on it rotates at the constant rate of 2.73 revolutions per second. Find the radial acceleration of a
Lostsunrise [7]

Answer:

110.9 m/s²

Explanation:

Given:

Distance of the tack from the rotational axis (r) = 37.7 cm

Constant rate of rotation (N) = 2.73 revolutions per second

Now, we know that,

1 revolution = 2\pi radians

So, 2.73 revolutions = 2.73\times 2\pi=17.153\ radians

Therefore, the angular velocity of the tack is, \omega=17.153\ rad/s

Now, radial acceleration of the tack is given as:

a_r=\omega^2 r

Plug in the given values and solve for a_r. This gives,

a_r=(17.153\ rad/s)^2\times 37.7\ cm\\a_r=294.225\times 37.7\ cm/s^2\\a_r=11092.28\ cm/s^2\\a_r=110.9\ m/s^2\ \ \ \ \ \ \ [1\ cm = 0.01\ m]

Therefore, the radial acceleration of the tack is 110.9 m/s².

4 0
3 years ago
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