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LiRa [457]
4 years ago
13

A space traveller leaves Earth for 10 years at .85c. According to an observer on Earth, how much time has passed?

Physics
1 answer:
eduard4 years ago
8 0
First of all, you didn't tell us WHO measured the "10 years".

If it was the people on Earth, then 10 years passed according to them.

If it was 10 years on the space traveler's clock,  then the clock in the
OTHER place, like on Earth, is subject to the relativistic 'time dilation'.

If the clocks are moving relative to each other, then the time interval measured
on either clock is equal to the interval measured on the other clock, divided by

       √(1 - v²/c²) .

You said that  v/c  = 0.85 .

v²/c² = (0.85)² = 0.7225

1 - v²/c² =  1 - 0.7225 = 0.2775

√(1 - v²/c²)  =  √0.2775 = 0.5268

If one clock counts up 10 years, then the other one counts up

(10years) / 0.5268 =  <em>18.983 years </em>


I believe that's the way to do this, and I'll gladly take your points,
but let me recommend that you get a second opinion before you
actually take off on your 10-year interstellar mission.

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NEED HELP ASAP !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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It would be c repeated processes-that are used in a variety of ways

5 0
3 years ago
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Answer:

B

Explanation:

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7 0
3 years ago
In need of some help here
Cloud [144]

Answer:

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Explanation:

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Alenkasestr [34]

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3 0
3 years ago
A single-turn plane loop of wire with a cross-sectional area 200 cm2 is perpendicular to a magnetic field that increases uniform
BartSMP [9]

Answer:

Induced current,I=2.87\times 10^{-3}\ A                                        

Explanation:

Given that,

Area of cross section of the wire, A=200\ cm^2=0.02\ m^2

Time, t = 2.2 s

Initial magnetic field, B_i=0.2\ T

Final magnetic field, B_f=2.8\ T

Resistance of the coil, R = 8 ohms

The expression for the induced emf is given by :

\epsilon=-\dfrac{d\phi}{dt}

\phi = magnetic flux

\epsilon=-\dfrac{d(BA)}{dt}

\epsilon=A\dfrac{d(B)}{dt}

\epsilon=A\dfrac{B_f-B_i}{t}

\epsilon=0.02\times \dfrac{2.8-0.2}{2.2}

\epsilon=-0.023\ volts

So, the induced emf in the loop is 0.023 volts. The induced current can be calculated using Ohm's law as :

\epsilon= IR

I=\dfrac{\epsilon}{R}

I=\dfrac{0.023}{8}

I=2.87\times 10^{-3}\ A

So, the magnitude of the induced current in the loop of wire is 2.87\times 10^{-3}\ A. Hence, this is the required solution.

7 0
3 years ago
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