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Kryger [21]
3 years ago
13

If the wavelength of the red line of hydrogen coming from a galaxy is 654.3 nanometers instead of the normal 656.3 nanometers, i

s the galaxy moving towards the Earth or away from the Earth, is it red shifted or blue shifted, and how fast is it moving
Physics
1 answer:
Delicious77 [7]3 years ago
5 0

Answer:

The Doppler effect says that, if the source of light is moving towards the viewer, then the light will turn to the blue side (the wavelength decreases) if the source of light is moving away, then the light will turn to the red side (the wavelength will increase)

In this case, we have that the light that normally has a wavelength of 654.3 nm, now has 656.3 nm. So we can see that the wavelength increased, which means that the galaxy is moving away from the Earth, is red shifted and the velocity can be found by the equation:

(λ - λ₀)/λ₀ = v/c

Here we have that:

λ = 656.3 nm

λ₀ = 654.3 nm

c = 3.10^6 m/s

and v is the velocity of the galaxy.

(the fact that the units are different does not matter because the units cancel with the divisions)

(656.3 nm - 654.3 nm)/654.3 nm = v/3*10^6m/s

0.003*3*10^6m/s = v = 9000 m/s

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Answer:

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Term: Monoatomic

Explanation:

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3 years ago
If a person tries to lift a heavy box for 5 seconds and can't make it budge,the work done on the box is less than zero
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Question not making a sence, Clarify what you wana ask

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3 years ago
The average distance of the planet mercury from the sun is 0.39 times the average distance of the earth from the sun. How long i
Stels [109]

Answer:

T_1=0.24y

Explanation:

Using Kepler's third law, we can relate the orbital periods of the planets and their average distances from the Sun, as follows:

(\frac{T_1}{T_2})^2=(\frac{D_1}{D_2})^3

Where T_1 and T_2 are the orbital periods of Mercury and Earth respectively. We have D_1=0.39D_2 and T_2=1y. Replacing this and solving for

T_1^2=T_2^2(\frac{D_1}{D_2})^3\\T_1^2=(1y)^2(\frac{0.39D_2}{D_2})^3\\T_1^2=1y^2(0.39)^3\\T_1^2=0.059319y^2\\T_1=0.24y

5 0
3 years ago
Please please help me, don't guess please my test is timed !!
dimulka [17.4K]

Answer:

a) J = F  t = 40 * .05 = 2 N-s

b)  J = 2 N-s     momentum changed by 2 N-s

c) Initial momentum appears to be zero

J = change in momentum = m v2 - m v1 = m v2 = 2 N-s

v2 = J / m = 2 / .057 = 35 m/s

d) if the impulse time was increased and the average force remained the same then the change in momentum would increase with a corresponding increase in velocity attained - note the increase in v2 in part c)

4 0
2 years ago
A circular rod with a gage length of 3.5 mm and a diameter of 2.8 cmcm is subjected to an axial load of 68 kN . If the modulus o
Crank

To solve this problem, we will apply the concepts related to the linear deformation of a body given by the relationship between the load applied over a given length, acting by the corresponding area unit and the modulus of elasticity. The mathematical representation of this is given as:

\delta = \frac{PL}{AE}

Where,

P = Axial Load

l = Gage length

A = Cross-sectional Area

E = Modulus of Elasticity

Our values are given as,

l = 3.5m

D = 0.028m

P = 68*10^3 N

E = 200GPa  

A = \frac{\pi}{4}(0.028)^2 \rightarrow 0.0006157m^2

Replacing we have,

\delta = \frac{PL}{AE}

\delta = \frac{( 68*10^3)(3.5)}{(0.0006157)(200*10^9)}

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\delta = 1.93mm

Therefore the change in length is 1.93mm

7 0
3 years ago
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