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Artemon [7]
3 years ago
9

Star #1 is approaching the Earth with speed v. Star #2 is receding from the Earth with the same speed v. Measurements of the sam

e spectral line from each of the stars show Doppler shifts in frequency. The light from which star will have the larger magnitude shift in frequency? a. star #1 b. Both stars will have the same shift. c. The value of the speed must be known before an answer can be found. d. star #2
Physics
1 answer:
leva [86]3 years ago
4 0

Answer:

b. Both stars will have the same shift.

Explanation:

It's a very simple problem to solve. Star 1 is approaching toward Earth with a speed v, so let's assume that the change in Doppler Shift is +F and Star 2 is moving away so the change in Doppler shift is -F. But it's time to notice the speed of both stars and that is same but only directions are different. speed is the main factor here. The magnitude of both shifts is F as we can see and + and - are showing there direction of motion. So, because of same amount of speed, both stars will have same shift magnitude. (Just the directions are different)

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Free Fall: A rock is thrown directly upward from the edge of a flat roof of a building that is 56.3 meters tall. The rock misses
Slav-nsk [51]

Answer:

v₀₁= 5.525 m / s

Explanation

Freefall Formulas :

The sign of acceleration due to gravity  (g) is positive if the object is going down and negative if the object is going up.

vf= v₀+gt  

vf²=v₀²+2*g*h

h= v₀t+ (1/2)*g*t²

Where:  

h: hight in meters (m)    

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

g: acceleration due to gravity in m/s²

Kinematics of the rock from the starting point with vo until it reaches its maximum height:

vf₁= v₀₁-gt₁  :vf₁ =0 to maximum height

0= v₀₁-gt₁

v₀₁ = g*t₁

t₁ =v₀₁ / g      Equation (1)

vf₁²= v₀₁²-2*g*h₁   : vf₁ =0 to maximum height

0 = v₀₁²-2*g*h₁

2*g*h₁ = v₀₁²

h₁ = (v₀₁²)/(2g)   Equation (2)

Kinematics of the rock when it falls from the maximum height until it touches the floor

h₂= v₀₂t+ (1/2)*g*t₂²  v₀₂=vf₁ =0

h₂= 0+ (1/2)*g*t₂²

h₂= (1/2)*g*t₂²   Equation (3)

Equation that relates h₁ to h₂

h₂=  h₁ + 56.3  ,  h₁ = (v₀₁²)/(2g)

h₂= (v₀₁²)/(2g) + 56.3  Equation (4)

Equation that relates t₁ to t₂

t₁ + t₂ =4 s

t₂ =4 -t₁

t₂ =4 -(v₀₁/g )

Calculation of v₀₁

We replace equation 4 and equation 5 in equation 3

(v₀₁²)/(2g) + 56.3 = (1/2)*g*(4 -(v₀₁/g ) )²

(v₀₁²)/(2g) + 56.3 = (1/2)*g* (16 - 2*4*(v₀₁/g )+((v₀₁/g )²)

we eliminate (v₀₁²)/(2g) on both sides of the equation

56.3 = (1/2)*g* (16 - 2*4*(v₀₁/g ))

56.3 = 78.4 - 4*v₀₁

4*v₀₁ =78.4-56.3

v₀₁= (78.4-56.3) / ( 4)

v₀₁= 5.525 m / s

7 0
3 years ago
Can an argon atom undergo vibrational motion?
azamat
No. the answer to the question if can an argon atom undergo vibrational motion is no. it can not even spin either. the argon atom, or the argon is a chemical element that is the third most abundant gas in the earth's atmosphere. it is ore than twice as abundance as water vapor. Thank you for this question.
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HELP ME! One airline limits the size of carry-on luggage to a volume of 40,000 cm^3. A passenger has a carry-on that has an area
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You must times the area by the volume, look at it as if the area is just one of 23 layers that makes up the volume.
1960x23=45080
so no it cannot be carried as it is 5080cm^3 over the limit
8 0
3 years ago
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Pani-rosa [81]
The total displacement is 4.0 m east.
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3 years ago
A positive object touches a neutral electroscope, and the laves separate. Then a negative object is brought near the electroscop
MaRussiya [10]

Answer:

If the positive object is touched to the electroscope, then the leaves would diverge because of the excess positive charge on the electroscope.

If then a negative object is brought near the electroscope, the excess positive charge on the leaves would be attracted to the external terminals of the electroscope and the leaves would again lose the distance of separation and be close to one another.

3 0
3 years ago
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