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____ [38]
3 years ago
12

Frame S' passes frame S in the usual way (positive directions). An object also moves in the positive direction. Which is true? T

he object's speed relative to S is greater than its speed relative to S'. The object's speed relative to S can be greater than or less than its speed relative to S', depending on the actual values. The object's speed relative to S is equal to its speed relative to S'. The object's speed relative to S is less than its speed relative to S'.
Physics
1 answer:
lisov135 [29]3 years ago
3 0

Answer:

The answer is "The object's speed relative to S can be greater than or less than its speed relative to S', depending on the actual values."

Explanation:

The S' frame and the object are moving in a positive direction. The object is moving with respect to the S frame so the S frame the rest frame

take the velocity of the object with respect to the rest frame as v and the velocity of the S' frame with respect S frame as v2

relative velocity of the object to the S' frame would be

Vrel = v2- v

This means the Vrel of the object with respect to the S' frame is less than the Vrel  of the object with respect to the S frame

However is the S' velocity is greater than that of the object then the Vrel of the object with respect to the S' frame is greater than the Vrel of the object with respect to the S frame.

This would mean the second option is the answer, the relative speed of the object depends on the actual values.

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Complete Question

You are performing a double slit experiment very similar to the one from DL by shining a laser on two nattow slits spaced 7.5 * 10^{-3} meters apart. However, by placing a piece of crystal in one of the slits, you are able to make it so that the rays of light that travel through the two slits are Ï out of phase with each other (that is to say, Ao,- ). If you observe that on a screen placed 4 meters from the two slits that the distance between the bright spot closest to center of the pattern is 1.5 cm, what is the wavelength of the laser?

Answer:

The  wavelength is  \lambda  =  56250 nm

Explanation:

From the question we are told that

   The  distance of slit separation is  d =  7.5 *10^{-3} \  m

   The  distance of the screen is  D =  4 \  m

    The  distance between the bright spot closest to the center of the interference  is  k   = 1.5 \ cm = 0.015 \  m

   

Generally the width of the central  maximum fringe produced is mathematically represented as

        y  =  2 *  k  = \frac{ D  *  \lambda}{d}

  =>    2 *  0.015 =  \frac{ \lambda  *  4}{ 7.5 *10^{-3}}

   =>   \lambda  =  56250 *10^{-9} \ m

=>      \lambda  =  56250 nm

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Answer:

240 ohms

Explanation:

From Ohms law we deduce that V=IR and making R the subject of the formula then R=V/I where R is resistance, I is current and V is coltage across. Substituting 120 V for V and 0.5 A for A then

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Learn more about conservation of energy here: brainly.com/question/166559

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