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The control panel on a spaceship contains a light that blinks every 2.00 s as observed by an astronaut in the ship. If the spaceship is moving past Earth with a speed of $0.750 c,$ determine (a) the proper time interval between blinks and (b) the time interval between blinks as observed by a person on Earth.

a) $t_{p}=2.005$

b) $\Delta t_{p}=3.02 \mathrm{sec}$

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Cornell University

Numerade Educator

University of Washington

University of Winnipeg

Well, Tom, mated by an astronaut in the spaceship. Uh, which is proper time is equal to two seconds. Uh, time, time majored time measured by. And this we're not. And astronaut in the ship in the ship. Uh, it's the proper time. Proper time is equal to will TP is the proper time. Uh, two seconds and speed of the spaceship. A speed off the spaceship. Speed off the spaceship is equal to we, and B is equal to, uh, 0.7 50 times the speed of light. Let's go party. Well, time it'd buy an astronaut in the spaceship is the proper time. And therefore, uh, time Mitchard by an astronaut in the spaceship. Appropriate time is equal to two point 00 seconds. Part B. The time interval between links is observed by a person on earth. We know that Delta T is equal to Delta TP, which is proper time. Dubai Dubai Square root door one minus. We squared, divided by C square. Let's split in the values when the proper time is equal to two seconds. Divided by scared Ruto one minus Well, we's 0.750 Therefore, we have Siegel 0.7 five seed or um, square Dimes C square, divided by C square. All right, and simplifying this we have for two divided by scared route off one minus 0.56 to 5, 56 25 And therefore delta t dine. It'd Time is equal to is equal to tree bind 02 seconds.