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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?

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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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

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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

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Answer:

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

Explanation:

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

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so here we know that

mass of boy = 70 kg

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now we know that

$F = ma$

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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

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

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

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Answer:

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.

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3 years ago