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Help me out with these 2 questions for 15 points.

Help me out with these 2 questions for 15 points.-example-1

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Explanation:

The time dilation formula is given by


F(t) = (t)/(√(1-v^2))

where t is the time measured by the moving observer and F(t) is the time measured by the stationary earth-bound observer and v is the velocity of the moving observer expressed as a fraction of the speed of light.

a) If the observer is moving at 80% of the speed of light and observes an event that lasts for 1 second, a stationary observer will see the same event occurring over a time period of


F(t) = \frac{1\:\text{s}}{√(1-(0.8)^2)}


\:\:\:\:\:\:\:=\frac{1\:\text{s}}{√(1-(0.8)^2)} =\frac{1\:\text{s}}{√(1-(0.64))}


\:\:\:\:\:\:\:=1.67\:\text{s}

This means that any event observed by this moving observer will be seen by a stationary observer to occur 67% longer.

b) Given:

t = 1 second

F(t) = 2 seconds

We need to find the speed of the observer such that an event seen by this observer will occur twice as long as seen by a stationary observer. Move the term containing the radical to the left side so the equation becomes


√(1-v^2) = (t)/(F(t))

Take the square of both sides, we get


1 - v^2 =(t^2)/(F^2(t))

Solving for v, we get


v^2 = 1 - (t^2)/(F^2(t))

or


v = \sqrt{1 - (t^2)/(F^2(t))}

Putting in the values for t and F(t) we get


v = \sqrt{1 - \frac{(1\:\text{s})^2}{(2\:\text{s})^2}}


v = \sqrt{1 - (1)/(4)} = √(0.75)


\:\:\:\:=0.866

This means that the observer must moves at 86.6% of the speed of light.

User JuneT
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