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kaheart [24]
3 years ago
15

You observe a distant galaxy. You find that a spectral line, resulting from an electron transition in hydrogen, is shifted from

its normal location in the visible part of the spectrum into the infrared part of the spectrum. What can you conclude?
Physics
1 answer:
alexandr1967 [171]3 years ago
6 0

Answer:

The galaxy is moving away from the observer

Explanation: when a galaxy is moving away from us, the light we percieve from it is "streched". Since the wavelength has an inverse raltionship whith frequency, the longer the wavelength is, the lower the frequency. And lower frequencies correspond to red and infrarred light.

So when we see the light has shifted to the infrarred part of the spectrum, it means the source is traveling away from us, making the light waves we percieve streched and move from visible light to infrarred.

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Structures on a bird feather act like a diffraction grating having 8000 8000 lines per centimeter. What is the angle of the firs
sammy [17]

Answer:

Angle of first order maximum, \theta=21.19^{\circ}

Explanation:

Given that,

Wavelength of the light, \lambda=452\ nm=452\times 10^{-9}\ m

Number of lines, N = 8000 per cm

The relation between the number of lines and the slit width is given by :

d=\dfrac{1}{N}

d=\dfrac{1}{8000/cm}  

d=0.000125\ cm=1.25\times 10^{-6}\ m

The equation of grating is given by :

d\ sin\theta=n\lambda

n = 1

sin\theta=\dfrac{\lambda}{d}

sin\theta=\dfrac{452\times 10^{-9}}{1.25\times 10^{-6}}

\theta=21.19^{\circ}

So, the angle of the first-order maximum is 21.19 degrees. Hence, this is the required solution.

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3 years ago
A 5.00 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the en
Brut [27]

Answer:

75 N

Explanation:

In this problem, the position of the crate at time t is given by

y(t)=2.80t+0.61t^3

The velocity of the crate vs time is given by the derivative of the position, so it is:

v(t)=y'(t)=\frac{d}{dt}(2.80t+0.61t^3)=2.80+1.83t^2

Similarly, the acceleration of the crate vs time is given by the derivative of the velocity, so it is:

a(t)=v'(t)=\frac{d}{dt}(2.80+1.83t^2)=3.66t [m/s^2]

According to Newton's second law of motion, the force acting on the crate is equal to the product between mass and acceleration, so:

F(t)=ma(t)

where

m = 5.00 kg is the mass of the crate

At t = 4.10 s, the acceleration of the crate is

a(4.10)=3.66\cdot 4.10 =15.0 m/s^2

And therefore, the force on the crate is:

F=ma=(5.00)(15.0)=75 N

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