Gravitation

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##### Christina K.

Rutgers, The State University of New Jersey

##### Marshall S.

University of Washington

##### Jared E.

University of Winnipeg

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### Video Transcript

in this problem. On the topic of relativity, we are given the mid lifetime for the mule on which is approximately 2.2 microseconds. We are then told to suppose that the neurons were very speed of 95% the speed of light or 0.95 c, and we are given the number of neurons. AT T is equal to zero. We want to find the observed lifetime of the neurons, and we want to know the number of neurons that remain after traveling a distance of the kilometers. Now at first, when the neurons are addressed, we know they have a mean lifetime of data TP, which is 2.2 microseconds. And if we look at a frame of reference, where the 0.95 c and we can calculate the time dilated mean lifetime of the neurons, which will call a tour and tour is equal to gamma that t So this is delta T divided by the square root of one minus the oversee old square. If we subject our values in here, this is too went to microseconds over mhm square root of one minus and V oversee is 0.95 and that's squared so calculating we get this to be seven microseconds. So in a frame in which the new ones are traveling at 95% the speed of light they observe lifetime will be seven microseconds. Next for Part B. We want to know how many millions will remain after traveling a distance of three kilometers. If the initial number is known now in a frame of reference where they travel at 0.95 c, the time required for them to travel three kilometers so called T and T is the distance divided by the speed. This is simply the distance of three kilometers, three times 10 to the three meters, divided by the speed which is 0.95 times the speed of light, which is three times 10 to the power, eight meters per second. So calculating we get this to be 1.5 times 10 to the minus five seconds, which is 10 0.5 micro cycles. Now if we know the number of started, a three kilometer trip and not which is five times 10, the power four, the number remaining at the end, which is n is equal to and not e the minus t over tall, which is the original number and not five times 10 to the power for multiplied. But he to the minus 10 0.5 microseconds, divided by the observed lifetime of seven microseconds, which gives us the number of new ones remaining after the 30 kilometre journey to be 1.1 times 10 to the power for.

University of Kwazulu-Natal

#### Topics

Gravitation

##### Christina K.

Rutgers, The State University of New Jersey

##### Marshall S.

University of Washington

##### Jared E.

University of Winnipeg

Lectures

Join Bootcamp