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Helen [10]
2 years ago
11

A thin beam of light enters a thick plastic sheet from air at an angle of 32.0° with the normal and continues in the sheet at an

angle of 23.0° with the normal. What is the index of refraction of the plastic?
Physics
1 answer:
Cloud [144]2 years ago
5 0

Answer:

1.36

Explanation:

n_{air} = Index of refraction of air = 1

n_{plastic} = Index of refraction of plastic = ?

i = angle of incidence in air = 32.0° deg

r = angle of refraction in plastic = 23.0° deg

Using Snell's law

n_{air} Sini = n_{plastic} Sinr

(1) SIn32.0 = n_{plastic} Sin23.0

n_{plastic} = 1.36

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Key concepts

Heart rate

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Introduction

As Valentine's Day approaches, we're increasingly confronted with "artistic" images of the heart. Real hearts hardly resemble to two-lobed shapes adorning cards and candy boxes this time of year. And the actual shape of the human heart is important for its function of supplying blood to the entire body. You have likely noticed that your heart beats more quickly when you exercise. But have you ever taken the time to observe how long it takes to return to its normal rate after you're done exercising? In this science activity you'll get to do some exercises to explore your own heart-rate recovery time.

Background

Your heart is continuously beating to keep blood circulating throughout your body. Its rate changes depending on your activity level; it is lower while you are asleep and at rest and higher while you exercise—to supply your muscles with enough freshly oxygenated blood to keep the functioning at a high level. Because your heart is also a muscle, exercise, in turn, helps keep it healthy. The American Heart Association recommends that a person does exercise that is vigorous enough to raise their heart rate to their target heart-rate zone—50 percent to 85 percent of their maximum heart rate, which is 220 beats per minute (bpm) minus their age for adults—for at least 30 minutes on most days, or about 150 minutes a week in total. So for a 20-year-old, the maximum heart rate would be 200 bpm, with a target heart-rate zone of 100 to 170 bpm. (For those 19 or younger, target zones can vary more than they do for adults.)

i think it will help you...if it help you ...please mark brainless

8 0
2 years ago
Suppose a 500 mb chart valid today at 12 Z indicates a large trough over the eastern US and a large ridge over the western US. A
Zinaida [17]

Answer:

An aircraft, flying in the vicinity of 18,000 ft altitude from west to east over the US at 12 Z today, will __LOSE___ altitude if the altimeter is not corrected

7 0
2 years ago
An automobile is traveling at a constant 15 m/s, then it undergoes acceleration from that moment forward. Which statement best d
creativ13 [48]

Answer:

d

Explanation:

7 0
2 years ago
Suppose that you have left a 200-mL cup of coffee sitting until it has cooled to 30∘C , which you find totally unacceptable. You
AleksAgata [21]

Answer:

\eta=0.5074\ or\ 50.74\%

Explanation:

<em><u>Considering the density & specific heat capacity of coffee to be equal to that of water.</u></em>

<em><u>GIVEN:</u></em>

  • density \rho=1\ g.mL^{-1}
  • specific heat c=4.186\ J.g^{-1}.K^{-1}
  • mass of coffee, m=200\times 1=200\ g
  • initial temperature of coffee, T_i=30^{\circ}C
  • final temperature of coffee, T_f=60^{\circ}C
  • power rating of oven, P=1100\ W
  • time taken to reach the final temperature, t=35\ s

<u>Heat released by the coffee to come to 60°C:</u>

Q=m.c.\Delta T

Q=200\times 4.186\times 30

Q=[tex]\eta=\frac{25116}{49500}\ J[/tex]

<u>Now the energy used by the oven in the given time:</u>

E=P.t

E=1100\times 45

E=49500\ J

Now the efficiency:

\eta=\frac{Q}{E}

\eta=0.5074\ or\ 50.74\%

8 0
2 years ago
An end of a light wire rod is bent into a hoop of radius r. The straight part of the rod has length l; a ball of mass M is attac
Roman55 [17]

Answer:

arcsin(\frac{R\mu}{(R+l)\sqrt{\mu^2+1}})

Explanation:

By the Law of Sines,

sin \theta = \frac{sin \phi R}{ l + R}

From Newton's Law,

mg = N\sqrt{\mu^2+1}

And the last equation again from Newton's Law,

\mu N = mgsin\phi

Then if we collect all equations together,

\mu N = mgsin\phi = N\sqrt{\mu^2+1}sin\phi\\

sin\theta = \frac{\mu R}{ (l + R)\sqrt{\mu^2+1}}

Thus,

\theta = arcsin(\frac{R\mu}{(R+l)\sqrt{\mu^2+1}})

8 0
2 years ago
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