Physics, 06.09.2019 23:10 noelmusulin1

In inertial frame s, two events occur at the same instant in time and 3 light- minutes apart in space. in inertial frame s", the same events occur at 5 light-minutes apart. what is the time interval between the events in s?

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Answer from: brebun4742

The time interval between the events in S its zero, and in S' its 4 min.

Explanation:

The space-time interval its defined as:

$( \Delta S )^ 2 \ = (c \ \Delta t)^2 \ - \ ( \Delta \vec{x})^2$

(of course, using the (+ - - -) metric signature).

In the inertial frame S, the events are separated by 3 light - minutes in space, and occur at the same instant in time, so:

$\Delta t \ = \ 0$

$| \ \Delta \vec{x} \ | \ = \ 3 \ min \ * \ c$

$( \Delta S )^ 2 \ = (c \ 0)^2 \ - \ ( 3 \ min \ * \ c)^2$

$( \Delta S )^ 2 \ = \ - \ ( 3 \ min \ * \ c)^2$

In the inertial frame S', the events are separated by 5 light - minutes in space, and we don't know how far are they separated in time, so:

$| \ \Delta \vec{x}' \ | \ = \ 5 \ min \ * \ c$ .

$( \Delta S' )^ 2 \ = \ (c \Delta t')^2 \ - \ ( 5 \ min \ * \ c)^2$

Now, between inertial frames the spacetime interval for the same events must be constant. So:

$\Delta S = \Delta S'$

$\ - \ ( 3 \ min \ * \ c)^2 = \ (c \ \Delta t')^2 \ - \ ( 5 \ min \ * \ c)^2$

$\ \ ( 5 \ min \ * \ c)^2 - \ ( 3 \ min \ * \ c)^2 = \ (c \ \Delta t')^2$

$25 \ min^2 \ c ^ 2 - 9 \ min^2 \ c ^2 \ = \ (c \ \Delta t')^2$

$16 \ min^2 \ c ^2 \ = \ (c \ \Delta t')^2$

$\sqrt{16 \ min^2 \ c ^2 \ } = c \ \Delta t'$

$4 \ min \ c \ = c \ \Delta t'$

$\Delta t' \ = 4 \ min$

So, in the S' inertial frame, the time interval its 4 min.

Answer from: Quest

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).

Answer from: Quest

Iam looking for this answer too!

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