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MrRa [10]
3 years ago
12

A graduate student has done a careful analysis of the spectrum of a star. While she has found lines from many elements, there wa

s not a trace of the element helium in the spectra she has been analyzing. From this she can now conclude:_______
Physics
1 answer:
disa [49]3 years ago
7 0

Explanation:

The lines in the spectra of star is due to absorption of light emitted by star by different elements  present in atmosphere of the star .

If some element is present in the outer atmosphere of he star , it will absorb a particular portion of light , for which there will be black line in the spectrum .

These are called Fraunhoffer's lines .

Now there is no such line at the place of Helium in the light spectrum . This means that Helium element is not present in the outer atmosphere of the star.

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An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi
Veronika [31]

Answer:

s=5.79\ km

\theta=47^{\circ} east of south

Explanation:

Given:

  • distance of the person form the initial position, d'=8.4\ km
  • direction of the person from the initial position, 47^{\circ} north of east
  • distance supposed to travel form the initial position, d=5.3\ km
  • direction supposed to travel from the initial position, due North

<u>Now refer the schematic for visualization of situation:</u>

y=d'.\sin47^{\circ}-d

y=8.4\times \sin47-5.3 ...............(1)

x=d'.\cos47^{\circ}

x=8.4\times \cos47^{\circ} .................(2)

<u>Now the direction of the desired position with respect to south:</u>

\tan\theta=\frac{y}{x}

\tan\theta=\frac{8.4\times \sin47}{8.4\times \cos47}

\theta=47^{\circ} east of south

<u>Now the distance from the current position to the desired position:</u>

s=\sqrt{x^2+y^2}

s=\sqrt{(8.4\times \cos47)^2+(8.4\times \sin47-5.3)^2}

s=5.79\ km

4 0
3 years ago
What is the wavelength of an AM radio wave in a vacuum if its frequency is 810 kHz?
dalvyx [7]
Answer- 370.114 meters
6 0
2 years ago
The stiffness of a particular spring is 42 N/m. One end of the spring is attached to a wall. A force of 2 N is required to hold
Liula [17]

Answer:

L_{o}=0.1224m

Explanation:

Given data

Force F=2 N

Length L=17 cm = 0.17 m

Spring Constant k=42 N/m

To find

Relaxed length of the spring

Solution

From Hooke's Law we know that

F_{spring}=k_{s}|s|\\F_{spring}=k_{s}(L-L_{o})\\ 2N=(42N/m)(0.17m-L_{o})\\2=7.14-42L_{o}\\-42L_{o}=2-7.14\\42L_{o}=5.14\\L_{o}=(5.14/42)\\L_{o}=0.1224m

6 0
3 years ago
Every object in the universe exerts a force on every other object. This force is called
MrMuchimi

Gravity is the correct answer.

6 0
3 years ago
I’m ugly ain’t I ??????
olga55 [171]
Aw don’t say that! I’m sure your beautiful:)
7 0
3 years ago
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