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melamori03 [73]
4 years ago
7

Two objects, Object A and Object B, need to be identified. Object A's index of refraction is determined to be 1.77, and Object B

's index of refraction is determined to be 1.333. Knowing this information, which of the following must be true? A. Light can pass through Object A faster than it can pass through Object B.
B. The optical density of Object A is lower than the optical density of Object B.
C. Light can pass through Object B faster than it can pass through Object A.
D. The optical density of Object B is higher than the optical density of Object A.
Physics
2 answers:
Slav-nsk [51]4 years ago
4 0

The correct answer is

C. Light can pass through Object B faster than it can pass through Object A.

In fact, the index of refraction of a material is defined as:

n=\frac{c}{v}

where c is the speed of light in vacuum and v is the speed of light in the material. Rearranging the equation, we can write the speed of light in the material as:

v=\frac{c}{n}

So we that, the smaller the refractive index n, the greater the speed of light in the material, v. In this problem, object B has lower refractive index than object A, so light travels faster in object B.

zlopas [31]4 years ago
3 0
<span>In the question "Two objects, Object A and Object B, need to be identified. Object A's index of refraction is determined to be 1.77, and Object B's index of refraction is determined to be 1.333. Knowing this information, which of the following must be true? A. Light can pass through Object A faster than it can pass through Object B." The correct answer is "Light can pass through Object B faster than it can pass through object A." (option C) The refractive index is a ratio of the speed of light in a medium relative to its speed in a vacuum. The refractive index of any other medium is defined relative to the refractive index of a vacuum, which is assigned a value of 1. Thus, a refractive index of 1.33 for water means that light travels 1.33 times faster in a vacuum than in water.</span>
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A satellite is in a circular Earth orbit of radius r. The area A enclosed by the orbit depends on r2 because A = πr2. Determine
Salsk061 [2.6K]

Answer:

a. T=r^{3/2}

b. K=\frac{1}{r}

c. v=\frac{1}{\sqrt{r}}

d. v=\sqrt{r}

Explanation:

To make analysis about the satellite circular earth the depends or r and A

T^2=\frac{4\pi}{GM}*r^3

K=\frac{GM*m}{2*r}

a.

T^2=r^3

T=r^{3/2}

b.

K=\frac{GM*m}{2}*\frac{1}{r}

K=\frac{1}{r}

c.

K=\frac{1}{2}*m*v^2

v^2=\frac{2*K}{m}=\frac{2*Gm*m}{2*m*r}

v=\frac{1}{\sqrt{r}}

d.

v=\sqrt{2*GMr}

v=\sqrt{r}

3 0
3 years ago
How do you convert seconds to minutes, seconds to hours , minutes to seconds and hours to seconds
Bess [88]

Answer:

You could memorize conversions, or use conversion charts, or do a quick internet search for help.

Explanation:

I'll do a few of the conversions for you

<u>Seconds to minutes: </u>

there are 60 seconds in one minute. So if you are wondering how many seconds are in 3 and 1/2 minutes, you would do this conversion:

60 sec/min <em>times </em>x sec/3.5 min

which can be written as 60 x 3.5 = 210. so "x" seconds would be 210 seconds in 3 1/2 minutes.

<u>you could do the same thing in the opposite direction for minutes to seconds:</u>

1 min has 60 seconds. So in 7.25 minutes, how many seconds are there?

1 min/60 sec <em>times </em>7.25 min/x sec

which can be written as 7.25 x 60 = 435. so "x" seconds would be 435 seconds in 7.25 minutes.

<u>hours to seconds:</u>

this one is slightly more complicated

In one hour there are 60 minutes, and in one minute there are 60 seconds.

so to convert from hours to seconds you would do this conversion:

1 hr/60 min times 1 min/60 sec. then the "min" would cancel out, and you would be left with the label "hr/sec". to do the math, it would be 1 hr / 60 x 60.

60x60 = 3600. so you would have 1 hr/3600 sec. So in one hour there are 3600 seconds.

so if you want to know how many seconds are in 6.75 hours:

6.75 hr/x sec <em>times </em>3600 sec/1 hr

6.75 x 3600 = 24,300 so there are 24,300 seconds in 6.75 hours.

I hope this helps :)

4 0
4 years ago
If a car used 260,000 W of power to complete a race in 15 s, how much work did the car do?
suter [353]

Answer: 3.9 MW

Explanation:

1 W = 1 J/s

260000 J/s (15 s) = 3,900,000 = 3.9 MW

3 0
3 years ago
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Mrac [35]

Answer:

a)α= 53.13°

b)The velocity at the highest point =  15 m/s

The acceleration at the highest point = 9.8m/s^2

c)h=15 m

V=18.02 m/s

Explanation:

Speed of water ,u= 25 m/s

So the horizontal component of speed u = u cos α

Given that horizontal distance cover by water in 3 s is 45 m.

So We know that in projectile motion horizontal acceleration is zero.

In horizontal direction

Distance = Velocity x time

45 =  u cos α  x 3

u cos α = 45

45 = 25 cos α x 3

 cos α = 45/75

α= 53.13°

So the velocity at the highest point =  u cos α

The velocity at the highest point =  15 m/s

The acceleration at the highest point = 9.8m/s^2

 Now the velocity along vertical direction(Vo) =  u sin α

      Vo= 25 sin 53.13°

Vo =20 m/s

h=V_o.t-\dfrac{1}{2}gt^2

h=20\times 3-\dfrac{1}{2}\times 10\times 3^2

h=15 m

So at 15 m above the ground water will strike .

The y-component of velocity after 3 sec

Vy= Vo - g t

Vy = 20 - 10 x 3

Vy= -10 m/s

The horizontal component of velocity will remain 15 m/s.

The resultant velocity

V=\sqrt{10^2+15^2}\ m/s

V=18.02 m/s

5 0
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shepuryov [24]

Answer:

548.8 J

Explanation:

5 0
3 years ago
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