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Amanda [17]
2 years ago
14

Which of Jafar's statements are correct regarding distance and displacement?

Physics
1 answer:
shepuryov [24]2 years ago
5 0

Answer:The distance and magnitude of displacement are sometimes equal." Jafar is correct. The distance traveled and the magnitude of displacement are equal if and only if the path is a straight line in one direction.

Explanation:

You might be interested in
Now consider θc, the angle at which the blue refracted ray hits the bottom surface of the diamond. If θc is larger than the crit
Dennis_Churaev [7]

Complete Question:

A beam of white light is incident on the surface of a diamond at an angle \theta_{a},  since the index of refraction depends on the light's wavelength, the different colors that comprise white light will spread out as they pass through the diamond. For example, the indices of refraction in diamond are n_{red} = 2.410 for red light and n_{blue} = 2.450 for blue light. Thus, blue light and red light are refracted at different angles inside the diamond. The surrounding air has n_{air} = 1.000.

Now consider θc, the angle at which the blue refracted ray hits the bottom surface of the diamond. If θc is larger than the critical angle θcrit, the light will not be refracted out into the air, but instead it will be totally internally reflected back into the diamond. Find θcrit. Express your answer in degrees to four significant figures.

Answer:

\theta_{crit} = 24.09^{0}

Explanation:

Only the blue refracted ray is related to the critical angle in this question

n_{air} = 1.000

n_{blue} = 2.450

The relationship between the critical angle(\theta_{crit}), n_{air} and n_{blue} can be given as sin \theta_{crit} = \frac{n_{air} }{n_{blue} }

sin \theta_{crit} = \frac{1 }{2.450 }\\\theta_{crit} = sin^{-1} \frac{1 }{2.450 }\\\theta_{crit} = sin^{-1} 0.4082^{0}\\  \theta_{crit} = 24.09^{0}

6 0
2 years ago
Hope you can answer this: A Student Visits A Farm And Makes These Notes In Her Journal.
Strike441 [17]

Answer:

Common Sense

Explanation:

The chick has probably seen other chicks get caught by the Hawk and knows not to go near, or saw a giant bird flying straight towards it and used common sense to identify it as danger and run away. Although if this is for a test or a grade or something, please do not use the answer, it is most likely incorrect. This is honestly my best answer.

8 0
2 years ago
The 20-lb cabinet is subjected to the force F = (3 + 2t) lb, where t is in seconds. If the cabinet is initially moving down the
Andre45 [30]

Answer:

t₁ = 0.95 s

Explanation:

In this chaos we must use the definition of Newton's second law

      F = m a = m dv / dt

      dv = F dt / m

Let's replace and integrate, let's take the upward direction of the plane as positive, the force is positive

       dv = ∫ (3 + 2t) dt / m

       v = (3 t + 2 t²/ 2) /m

Let's evaluate between the lower limit t = 0 v = -6 ft / s (going down) to the upper limit   t = t and v = 0

       0 - (-6) = (3 (t- 0) + (t² -0)) / m

       t² + 3t -6m = 0

Let's look for the mass

      W = mg

      m = W / g

      m = 20/32

      m = 0.625 slug

Let's solve the second degree equation

     t² + 3t -3.75 = 0

     t = (-3 ± √ (32 + 4 1 3.75)) / 2

     t = (-3 ± 4,899) / 2

     t₁ = 0.95 s

     t₂ = -3.95 s

We take the positive time

6 0
2 years ago
A ray of yellow light ( f = 5.09 × 1014 hz) travels at a speed of 2.04 × 108 meters per second in
SashulF [63]
Velocity = fλ

where f is frequency in Hz, and λ is wavelength in meters.

<span>2.04 * 10⁸ m/s =  5.09 * 10¹⁴  Hz   *  λ </span>

<span>(2.04 * 10⁸ m/s) / (5.09 * 10¹⁴  Hz ) = λ </span>

<span>4.007*10⁻⁷  m =  λ </span>

<span>The wavelength of the yellow light = 4.007*10⁻⁷  m<span> </span></span>
6 0
2 years ago
To determine the y-component of a projectile’s velocity, what operation is performed on the angle of the launch?
koban [17]

<em>To determine the y component of velocity of a projectile </em><u><em>sine </em></u><em>operation is performed on the angle of launch.</em>

<u>Answer:</u> <em>sine</em>

<u>Explanation:</u>

Thus a_x=0,a_y=g

The initial velocity u can be resolved along two directions.

Along the X direction initial velocity = u cos θ

Along y direction initial velocity= u sin θ

From the equation of motion v= u+at

Thus velocity along x direction v_x=u cos θ

Velocity along y direction v_y= u sinθ -gt

Sign of g is negative.

3 0
2 years ago
Read 2 more answers
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