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azamat
1 year ago
6

in comparing the galilean transformations with the lorentz transformations, what multiplicative factor appears in the lorentz tr

ansformations that is not in the galilean ones?
Mathematics
1 answer:
raketka [301]1 year ago
7 0

In Lorentz transformations we have velocity based factors that is not present in Galilean ones.

<h3>What is lorentz transformations?</h3>

The relationship between two distinct coordinate frames that are moving relative to one another at a constant speed is known as a Lorentz transformation. Dutch physicist Hendrik Lorentz is credited with coining the name of the transformation. There are two frames of reference: Inertial Frames, which refer to motion that has a constant speed.

<h3>What are Galilean transformations?</h3>

The relationship between two distinct coordinate frames that are moving relative to one another at a constant speed is known as a Lorentz transformation. Dutch scientist Hendrik Lorentz is credited with coining the term transformation. There are two frames of reference: Inertial Frames, which relate to motion that has a constant speed.

Any and all rulers and other self-contained length standards, such as those you might use to set up a measurement frame, can be contracted in length based on velocity. This results in the Lorentz factor at the place in the LT.

Any and all clocks and other physically independent systems, such as those used to record events in a measurement frame, are subject to velocity time dilation. As a result, the LF is temporarily in the LT.

The fact that applying Einstein synchronisation individually for each measurement frame turns out to be natural. Consequently, the position-dependent synchronisation offset between the clocks of one measurement frame and those of the other is represented by the enigmatic second term in the LT for the time.

To learn more about transformations click the following link :-

brainly.com/question/1548871

#SPJ4

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Step-by-step explanation:

Let y(x)=f(x)g(x)h(x) where

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Recall that the derivative of the product of functions is

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so taking the derivatives of the individual functions, we get

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So the derivative of y(x) is given by

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\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:- 60x^8(1 -2x^5)^5(5 - 8x^3)^2

\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:- 48x^6(1 -2x^5)^6 2(5 - 8x^3)

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\large{ \tt{❃ \: EXPLANATION}} :

  • We're provide - Sin θ = \frac{12}{37} which means 12 is the perpendicular & 37 is the hypotenuse [ Since Sin θ = \tt{  \frac{p}{h}} ] . We're asked to find out tan θ ].

\large{ \tt{❁ \: USING \: PYTHAGORAS \: THEOREM}} :

\large{ \tt{❊ \:  {h}^{2}  =  {p}^{2}  +  {b}^{2} }}

\large{ \tt{⇢ {p}^{2} +  {b}^{2}   =  {h}^{2} }}

\large{ \tt{⇢ \:  {b}^{2}  =  {h}^{2}  -  {p}^{2} }}

\large{ \tt{⇢ \:  {b}^{2} = {37}^{2} -  {12}^{2}   }}

\large{ \tt{⇢ \:  {b}^{2}  = 1369 - 144}}

\large{ \tt{ ⇢{b}^{2}  = 1225}}

\large{ \tt{⇢ \: b =  \sqrt{1225}}}

\large{ \tt{⇢ \: b = 35  \: \text {units}}}

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\large{ \tt{➝ \: Tan  \: \theta =  \frac{p}{b}  = \boxed{ \tt{  \frac{12}{35} }}}}

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