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Alenkasestr [34]
3 years ago
5

From laboratory measurements, we know that a particular spectral line formed by hydrogen appears at a wavelength of 486.1 nanome

ters (nm). The spectrum of a particular star shows the same hydrogen line appearing at a wavelength of 485.9 nm. What can we conclude?
- The "star" actually is a planet.
- The star is moving away from us.
- The star is getting colder.
- The star is getting hotter.
- The star is moving toward us.
Physics
1 answer:
aksik [14]3 years ago
6 0

Answer:

The star is moving toward us

Explanation:

The wavelength of a distant object changes due to the change in the distance between the observer and the object. This is known as the Doppler effect.

If the wavelength decreases this means that the wavelength in going towards blue which is shorter wavelength. This is known as blue shift. If blue shift occurs then it means that the object is coming closer to the observer.

Hence, the star described here has blue shifted and is moving closer to us.

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Sati [7]

Answer:

that answer is

b. 1/4,000

8 0
3 years ago
The stars, Rigel and Betelgeuse, are both found in the constellation Orion. Rigel is a blue supergiant, and Betelgeuse is a red
SIZIF [17.4K]

Answer:

batrix

Explanation:

6 0
3 years ago
Coherent light of wavelength 525 nm passes through two thin slits that are 4.15×10^(−2) mm apart and then falls on a screen 7
IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

sin \theta=\frac{n \lambda}{d}

where

n is the order of the maximum

\lambda is the wavelength

a is the distance between the slits

In this problem,

n = 5

\lambda=525 nm =5.25\cdot 10^{-7} m

a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m

So we find

\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}

And given the distance of the screen from the slits,

D=75.0 cm = 0.75 m

The distance of the 5th  bright fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}

And the distance of the 8th dark fringe from the central bright fringe will be given by

y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm

5 0
3 years ago
A 12 kg box is pulled across the floor with a 48 N horizontal force. If the force of friction is 12 N, what is the acceleration
Oksi-84 [34.3K]

Answer:

The acceleration of the box is 3 m/s²

Explanation:

Given;

mass of the box, m = 12 kg

horizontal force pulling the box forward, Fx = 48 N

frictional force acting against the box in opposite direction, Fk = 12 N

The net horizontal force on the box, F = 48 N - 12 N

The net horizontal force on the box, F = 36 N

Apply Newton's second law of motion to determine the acceleration of the box;

F = ma

where;

F is the net horizontal force on the box

a is the acceleration of the box

a = F / m

a = 36 / 12

a = 3 m/s²

Therefore, the acceleration of the box is 3 m/s²

7 0
3 years ago
GIVING 94 POINTS!!!!!! PLS HELP MEH
pishuonlain [190]

Answer:

A) asteroid

Explanation:

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3 years ago
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