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An interstellar space probe is launched from Earth. After a brief period of acceleration, it moves with a constant velocity, 70.0% of the speed of light. Its nuclear - powered batteries supply the energy to keep its data transmitter active continuously. The batteries have a lifetime of 15.0 years as measured in a rest frame. (a) How long do the batteries on the space probe last as measured by mission control on Earth? (b) How far is the probe from Earth when its batteries fail as measured by mission control? (c) How far is the probe from Earth as measured by its built - in trip odometer when its batteries fail? (d) For what total time after launch are data received from the probe by mission control? Note that radio waves travel at the speed of light and fill the space between the probe and Earth at the time the battery fails.

a) 21.0$y r$

b) 14.7$l y$

c) 10.5$l y$

d) 35.7$y r$

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well using time dilation equation. Delta T is equal to delta. T p divided by the square root off one minus. We squared, divided by C square. Well, he appropriate time is, uh, 15 years divided by square root off one minus. We leave 70% which is, uh, 0.700 all square. See, school gets canceled with C square, and therefore Delta T is equal to 21 point, uh, zero years. Well, this is part of it. Now, let's to part B. Well, using our disk unsteady is equal to speed multiplied by time. Well, speeding this case is 0.770 times the speed of light multiplied by, uh, T is a 21 point, uh, zero years and this is equal to a 14 point are seven light years. So distances, 14.7 light years. Uh, bark. See, uh, here, using a deep prime is equal to wheat multiplied with proper time. And therefore, D prime is well, d prime physical to multiply delta T p. And the 0.7 00 times the speed of light multiplied by proper time is 15 years. And thus D prime is equal to 10.5 light years. Okay, No party Well in body minutes. Calkins don't all ears us. A total years is, uh, to 21.0 years was, uh, 14.7 here, which gives us a 35.7 years.