In Example 2.5, we noted that Anna could go wherever she wished in as little time as desired by going fast enough to length−contract the distance to an arbitrarily small value. This overlooks a physiological limitation. Accelerations greater than about 30g are fatal, and there are serious concerns about the effects of prolonged accelerations greater than 1g. Here we see how far a person could go under a constant acceleration of 1g, producing a comfortable artificial gravity. (a) Though traveler Anna accelerates, Bob, being on near−inertial Earth, is a reliable observer and will see less time go by on Anna’s clock ( d t ' ) than on his own ( d t ) . Thus, d t ' = ( 1 / γ u ) d t , where u isAnna’s instantaneous speed relative to Bob, Using the result of Exercise 117(c), with g replacing F / m , substitute for u , then integrate to show that t = c g sinh g t ' c (b) How much time goes by for observes on Earth as they “see” Anna age 20 years? (c) Using the result of Exercise 119, show that when Anna has aged a time t ' . She is a distance fromEarth (according to Earth observers) of x = c 2 g ( cosh g t ' c − 1 ) (d) If Anna accelerates away from Earth while aging 20 years and then slows to a stop while aging another 20, how far away from Earth will she end up, and how much time will have passed on Earth?
In Example 2.5, we noted that Anna could go wherever she wished in as little time as desired by going fast enough to length−contract the distance to an arbitrarily small value. This overlooks a physiological limitation. Accelerations greater than about 30g are fatal, and there are serious concerns about the effects of prolonged accelerations greater than 1g. Here we see how far a person could go under a constant acceleration of 1g, producing a comfortable artificial gravity. (a) Though traveler Anna accelerates, Bob, being on near−inertial Earth, is a reliable observer and will see less time go by on Anna’s clock ( d t ' ) than on his own ( d t ) . Thus, d t ' = ( 1 / γ u ) d t , where u isAnna’s instantaneous speed relative to Bob, Using the result of Exercise 117(c), with g replacing F / m , substitute for u , then integrate to show that t = c g sinh g t ' c (b) How much time goes by for observes on Earth as they “see” Anna age 20 years? (c) Using the result of Exercise 119, show that when Anna has aged a time t ' . She is a distance fromEarth (according to Earth observers) of x = c 2 g ( cosh g t ' c − 1 ) (d) If Anna accelerates away from Earth while aging 20 years and then slows to a stop while aging another 20, how far away from Earth will she end up, and how much time will have passed on Earth?
In Example 2.5, we noted that Anna could go wherever she wished in as little time as desired by going fast enough to length−contract the distance to an arbitrarily small value. This overlooks a physiological limitation. Accelerations greater than about 30g are fatal, and there are serious concerns about the effects of prolonged accelerations greater than 1g. Here we see how far a person could go under a constant acceleration of 1g, producing a comfortable artificial gravity. (a) Though traveler Anna accelerates, Bob, being on near−inertial Earth, is a reliable observer and will see less time go by on Anna’s clock
(
d
t
'
)
than on his own
(
d
t
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. Thus,
d
t
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=
(
1
/
γ
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)
d
t
, where u isAnna’s instantaneous speed relative to Bob, Using the result of Exercise 117(c), with g replacing
F
/
m
, substitute for u, then integrate to show that
t
=
c
g
sinh
g
t
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c
(b) How much time goes by for observes on Earth as they “see” Anna age 20 years? (c) Using the result of Exercise 119, show that when Anna has aged a time
t
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. She is a distance fromEarth (according to Earth observers) of
x
=
c
2
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(
cosh
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−
1
)
(d) If Anna accelerates away from Earth while aging 20 years and then slows to a stop while aging another 20, how far away from Earth will she end up, and how much time will have passed on Earth?
Please solve and answer this problem correctly please. Thank you!!
You're on an interplanetary mission, in an orbit around the Sun. Suppose you make a maneuver that brings your perihelion in closer to the Sun but leaves your aphelion unchanged. Then you must have
Question 2 options:
sped up at perihelion
sped up at aphelion
slowed down at perihelion
slowed down at aphelion
The force of the quadriceps (Fq) and force of the patellar tendon (Fp) is identical (i.e., 1000 N each). In the figure below angle in blue is Θ and the in green is half Θ (i.e., Θ/2). A) Calculate the patellar reaction force (i.e., R resultant vector is the sum of the horizontal component of the quadriceps and patellar tendon force) at the following joint angles: you need to provide a diagram showing the vector and its components for each part. a1) Θ = 160 degrees, a2) Θ = 90 degrees. NOTE: USE ONLY TRIGNOMETRIC FUNCTIONS (SIN/TAN/COS, NO LAW OF COSINES, NO COMPLICATED ALGEBRAIC EQUATIONS OR ANYTHING ELSE, ETC. Question A has 2 parts!
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