Some distant galaxies are moving away from us at speeds greater than 0.5 c . What is the speed of the light received on Earth from these galaxies? Explain.
Some distant galaxies are moving away from us at speeds greater than 0.5 c . What is the speed of the light received on Earth from these galaxies? Explain.
Some distant galaxies are moving away from us at speeds greater than 0.5c. What is the speed of the light received on Earth from these galaxies? Explain.
Definition Definition Rate at which light travels, measured in a vacuum. The speed of light is a universal physical constant used in many areas of physics, most commonly denoted by the letter c . The value of the speed of light c = 299,792,458 m/s, but for most of the calculations, the value of the speed of light is approximated as c = 3 x 10 8 m/s.
Expert Solution & Answer
To determine
The speed of light received on Earth from distant galaxies.
Answer to Problem 1CQ
The speed of light received on Earth from distant galaxies is
c.
Explanation of Solution
According to second postulate of relativity, the light travels with same speed, regardless of whether the source or the observer is in motion or not.
The distant galaxies are moving away from us at speeds greater than
0.5c. The light received on Earth coming from these galaxies will be received at the speed of light in accordance with the second postulate of relativity.
The motion of galaxies relative to Earth will not have any effect on the speed of light coming from these galaxies according to the second postulate of relativity.
Conclusion:
Therefore, the speed of light received on Earth from distant galaxies is
c.
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12. A stone is dropped from the top of a cliff. It is seen to hit the ground below
after 3.55 s. How high is the cliff?
13. A ball is dropped from rest at the top of a building that is 320 m tall. Assuming
no air resistance, what is the speed of the ball just before it strikes the ground?
14. Estimate (a) how long it took King Kong to fall straight down from the top
of the Empire State Building (280m high), and (b) his velocity just before
"landing".
Useful equations
For Constant Velocity:
V =>
D
X = V₁t + Xo
For Constant Acceleration:
Vr = V + at
X = Xo+Vot +
v=V+2a(X-Xo)
\prom = V +V
V velocity
t = time
D Distance
X = Final Position
Xo Initial Position
V = Final Velocity
Vo Initial Velocity
a = acceleration
For free fall
Yf
= Final Position
Yo Initial Position
g = 9.80
m
$2
For free fall:
V = V + gt
Y=Yo+Vo t +
+gt
V,² = V₁²+2g (Y-Yo)
V+Vo
Vprom=
2
6
Solve the problems
A 11 kg weight is attached to a spring with constant k = 99 N/m and subjected to an external force
F(t) =-704 sin(5t). The weight is initially displaced 4 meters above equilibrium and given an
upward velocity of 5 m/s. Find its displacement for t> 0.
y(t)
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Human Physiology: An Integrated Approach (8th Edition)
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