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VladimirAG [237]
3 years ago
7

n hydrogen, the transition from level 1 to level 2 has a rest wavelength of 121.6 nm. Suppose you see this line at a wavelength

of 124.2 nm in a Star. What is the velocity of this in the direction of the Earth?
Physics
1 answer:
Irina-Kira [14]3 years ago
7 0

Answer:

velocity of this in the direction of the Earth is 6.4 × 10^{6}  m/s

Explanation:

given data

wavelength \lambda 1 = 121.6 nm

wavelength \lambda 2 = 124.2 nm

solution

we get here first change in wavelength that is

\triangle \lambda = 124.2 nm - 121.6 nm    ...............1

\triangle \lambda = 2.6 nm  

we get here velocity of direction of the Earth that is express as

\frac{\triangle \lambda }{\lambda 1} =\frac{v}{c}     ............2

put here value we get v

v = \frac{3\times 10^8\times 2.6}{121.6}    

v = 6.4 × 10^{6}  m/s

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Crank

Answer: 350 ms

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3 years ago
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A large fake cookie sliding on a horizontal surface is attached to one end of a horizontal spring with spring constant k = 440 N
irinina [24]

Answer:

a) 0.275 m b) 13.6 J

Explanation:

In absence of friction, the energy is exchanged between the spring (potential energy) and the cookie (kinetic energy), so at any point, the sum of both energies must be the same:

E = ½ kx2 + ½ mv2

If we take as initial state, the instant when the cookie is passing through the spring’s equilibrium position, all the energy is kinetic, and we know that is equal to 20.0 J.

After sliding to the right, while is being acted on by a friction force, it came momentarily at rest. At this point, the initial kinetic energy, has become potential elastic energy, in part, and in thermal energy also, represented by the work done by the friction force.

So, for this state, we can say the following:

Ki = Uf + Eth = ½* k*d2 + Ff*d

20.0J = ½ *440 N/m* d2 + 11.0 *d, where d is the compressed length of the spring, which is equal to the distance travelled by the cookie before coming momentarily at rest.

We have a quadratic equation, that, after simplifying terms, can be solved as follows, applying the quadratic formula:

d = -0.05/2 +/- √0.090625 = -0.025 +/- 0.3 = 0.275 m (we take the positive root)

b) If we take as our new initial status the moment at which the spring is compressed, and the cookie is at rest, all the energy is potential:

E = Ui = 1/2 k d²

In this case, d is the same value that we got in a), i.e., 0.275 m (as the distance travelled by the cookie after going through the equilibrium point is the same length that the spring have been compressed).

E= 1/2 440 N/m . (0.275)m² = 16.6 J

When the cookie passes again through the equilibrium position, the energy will be in part kinetic, and in part, it will have become thermal energy again.

So, we can write the following equation:

Kf = Ui - Ff.d = 16.6 J - 11.0 (0.275) m = 16.6 J - 3.03 J = 13.6 J

3 0
3 years ago
The International Space Station has a mass of 1.8 × 105 kg. A 70.0-kg astronaut inside the station pushes off one wall of the st
Aleonysh [2.5K]

Answer:

a = 5.83 \times 10^{-4} m/s^2

Explanation:

Since the system is in international space station

so here we can say that net force on the system is zero here

so Force by the astronaut on the space station = Force due to space station on boy

so here we know that

mass of boy = 70 kg

acceleration of boy = 1.50 m/s^2

now we know that

F = ma

F = 70(1.50) = 105 N

now for the space station will be same as above force

F = ma

105 = 1.8 \times 10^5 (a)

a = \frac{105}{1.8 \times 10^5}

a = 5.83 \times 10^{-4} m/s^2

3 0
3 years ago
A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.
MakcuM [25]

Answer:

b. 9.5°C

Explanation:

m_i = Mass of ice = 50 g

T_i = Initial temperature of water and Aluminum = 30°C

L_f = Latent heat of fusion = 3.33\times 10^5\ J/kg^{\circ}C

m_w = Mass of water = 200 g

c_w = Specific heat of water = 4186 J/kg⋅°C

m_{Al} = Mass of Aluminum = 80 g

c_{Al} = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C

The final equilibrium temperature is 9.50022°C

4 0
3 years ago
How large is the Milky Way Galaxy? A. It is the largest galaxy ever observed. B. It takes up over half of the known universe. C.
lisabon 2012 [21]

Answer:

D. It is very small when compared to the universe

Explanation:

The Milky Way can be regarded galaxy which has Solar System in it. Milky way gives the description of appearance of galaxy from Earth, it is a hazy band of light that's been formed from the stars which can be visualized in the sky during the night, though it cannot be sorted by mere human eyes. Milky Way has existed for about 13.51 billion years with the radius of 52,850 light years. the Number of stars in milky way is about 100-400 billion. It should be noted that themilky way galaxy is a very large galaxy but It is very small when compared to the universe

.

4 0
3 years ago
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