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melamori03 [73]
3 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]3 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]3 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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Answer:

In space we feel weightlessness because the earth's gravity has less effect on us. The Earth's gravitational attraction at those altitudes is only about 11% less than it is at the Earth's surface. If you had a ladder that could reach as high as the shuttle's orbit, your weight would be 11% less at the top.

Explanation:

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A space probe produces a radio signal pulse. If the pulse reaches Earth 12.3 seconds after it is emitted by the probe, what is t
adoni [48]

Answer:

Distance = 3.69 × 10^9 m

The distance from the probe to Earth is 3.69 × 10^9 m

Explanation:

Distance from the probe to the Earth can be derived using the simple motion formula;

Distance = speed × time .....1

Since a radio signal uses an electromagnetic wave to transfer signal, it has the same speed as the speed of light.

Speed of radio signal = speed of light = 3.0 × 10^8 m/s

time taken to reach the earth = 12.3 seconds

Substituting the values of speed and time into equation 1;

Distance = 3.0 × 10^8 m/s × 12.3 s

Distance = 36.9 × 10^8 m

Distance = 3.69 × 10^9 m

Note: all electromagnetic radiation have the same speed which is equal to 3.0 × 10^8 m/s

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2 years ago
A bike, a truck, and a train—all without passengers, motors, or engines—roll down the same hill. Put the vehicles in order from
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3 years ago
A block is given a very brief push up a 20.0 degree frictionless incline to give it an initial speed of 12.0 m/s.(a) How far alo
Orlov [11]

Explanation:

(a)   Net force acting on the block is as follows.

           F_{net} = -mg Sin (\theta)

or,           ma = -mg Sin (\theta)[/tex]

                 a = -g Sin (\theta)

                    = -9.8 \times Sin (20^{o})

                    = -3.35 m/s^{2}

According to the kinematic equation of motion,

             v^{2} - v^{2}_{o} = 2as

Distance traveled by the block before stopping is as follows.

     s = \frac{v^{2} - v^{2}_{o}}{2a}

        = \frac{(0)^{2} - (12.0)^{2}_{o}}{2 \times -3.35}

        = 21.5 m

According to the kinematic equation of motion,

               v = v_{o} + at

      0 = 12.0 m/s + \frac{1}{2} \times -3.35 m/s^{2} \times t

   t_{1} = 7.16 sec

Therefore, before coming to rest the surface of the plane will slide the box till 7.16 sec.

(b)    When the block is moving down the inline then net force acting on the block is as follows.

                 F_{net} = -mg Sin (\theta)

                ma = mg Sin (\theta)

                    a = g Sin (\theta)

                       = 9.8 m/s^{2} \times Sin (20^{o})

                       = 3.35 m/s^{2}

Kinematics equation of the motion is as follows.

                   s = v_{o}t + \frac{1}{2}at^{2}

      21.5 m = 0 + \frac{1}{2} \times 3.35 m/s^{2} \times t^{2}

     t_{2} = \sqrt{\frac{2 \times 21.5 m}{3.35 m/s^{2}}}

             = 3.58 sec

Hence, total time taken by the block to return to its starting position is as follows.

               t = t_{1} + t_{2}

                 = 7.16 sec + 3.58 sec

                 = 10.7 sec

Thus, we can conclude that 10.7 sec time it take to return to its starting position.

3 0
3 years ago
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