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Scorpion4ik [409]
3 years ago
6

The Sun’s surface temperature is about 5800 K and its spectrum peaks at 5000 Å. An O-type star’s surface temperature may be 40,0

00 K. (a) According to Wien’s law, at what wavelength does its spectrum peak? (b) In what part of the spectrum might that peak be? (c) Can the peak be observed with the Keck telescopes on Maunakea in Hawaii? Explai
Physics
1 answer:
nirvana33 [79]3 years ago
7 0

(a) 7.25\cdot 10^{-8}m

Wien's displacement law is summarized by the equation

\lambda = \frac{b}{T}

where

\lambda is the peak wavelength

b=2.898\cdot 10^{-3} m \cdot K is Wien's displacement constant

T is the absolute temperature at the surface of the star

For an O-type star, we have

T = 40,000 K

Therefore, its peak wavelength is

\lambda = \frac{2.898\cdot 10^{-3}}{40000}=7.25\cdot 10^{-8}m

(b) Ultraviolet

We can answer this part by looking at the wavelength range of the different parts of the electromagnetic spectrum:

gamma rays  

X-rays  1 nm - 1 pm

ultraviolet  380 nm - 1 nm

visible light  750 nm - 380 nm

infrared  25 \mu m - 750 nm

microwaves  1 mm - 25 \mu m

radio waves  > 1 mm

The peak wavelength of this star is

\lambda=7.25\cdot 10^{-8}m=72.5 nm

Therefore, it falls in the ultraviolet region.

(c) No

The Keck telescopes is actually a system of 2 telescopes in the Keck Observatory, located in Mauna kea, Hawai.

The two telescopes, thanks to several instruments, are able to detect  much of the electromagnetic radiation in the visible ligth and infrared parts of the spectrum. However, they are not able to detect light in the ultraviolet region: therefore, they cannot observe the star mentioned in the previous part of the problem.

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In which situation is work being done?
Keith_Richards [23]

Answer: D. The force and displacement are in the same direction.

Explanation:

The Work W done by a Force F refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path with distance d.  

Work is a scalar magnitude, and its unit in the International System of Units is the Joule (like energy).

Now, when the applied force is constant and the direction of the force and the direction of the displacement are <u>parallel</u>, the equation to calculate it is:  

W=(F)(d) (1)

When they are not parallel, both directions form an angle, let's call it \alpha. In that case the expression to calculate the Work is:  

W=Fdcos{\alpha} (2)

When the force and displacement are perpendicular to each other,\alpha=90\° and <u>no work is done</u>.

4 0
3 years ago
Would the solar panel work under a fluorescent or halogen light? explain your response being sure to relate your observations to
lina2011 [118]

A solar panel are most efficient under natural sunlight, however, a solar panel can work using artificial light simply because a solar panel collects photons which collide with silicon atoms transferring their energy which cause them to lose electrons.

3 0
3 years ago
A vertical spring stretches 4.0 cm when a 12-g object is hung from it. The object is replaced with a block of mass 28 g that osc
marin [14]

Answer:

Time period of the osculation will be 0.0671 sec  

Explanation:

It is given a vertical spring is stretched by 4 cm

So change in length of the spring x = 4 cm = 0.04 m

Mass which is hung from it m = 12 gram = 0.012 kg

Sprig force will be equal to weight of the mass

So kx=mg

k\times 0.04=0.012\times 9.8

k = 244.7 N/m

Now new mass is m = 28 gram = 0.028 kg

So time period with new mass will be

T=2\pi \sqrt{\frac{m}{k}}

=2\times 3.14 \sqrt{\frac{0.028}{244.7}}=0.0671sec

4 0
3 years ago
The correct formula for copper (I) bromide is —
Ad libitum [116K]

Answer:

The correct formula is the first one.

Explanation:

Copper has two valences I and II, in this example, it's mentioned that copper's valence is I.

Then we have to look in all the formulas for one formula where copper has that valence, and that valence is in formula 1 and 3. Number 4 is discarted that formula is incorrect, that formula doesn't exist.

Number 2 is also discarted because in this formula Cu has a valence of 2.

Number three is discarted because here Bromide has a valence of 2 and that is incorrect, Bromide's valence is 1.

8 0
3 years ago
Using the superposition method, calculate the current through R5 in Figure 8-71
Natasha2012 [34]

by superposition method we can find current in R5

here first let say only 2V battery is present in the circuit

now the equivalent resistance to be found for which we can say

2.2 k ohm and 1 k ohm is connected in parallel

r_1 = \frac{2.2 * 1}{2.2 + 1}

r_1 = 0.6875 k ohm

now it is in series with 1 k ohm and then that part is in parallel with 2.2 k ohm

r_2 = \frac{2.2* (1+0.6875)}{2.2 + (1+0.6875)}

r_2 = 0.95 k ohm

now the current flowing through the battery is

i = \frac{2}{1 + 0.95} = 1.02 mA

now this will divide into R3 and R2 so current flowing in R3 will be

i_1 = \frac{2.2}{2.2+1.6875}*1.02 = 0.58 mA

now this will again divide in R4 and R5

so current in R5 will be

i_5 = \frac{R_4}{R_4 + R_5}* i_1

i_5 = 0.18 mA

now when only 3 V battery is present in the circuit

R1 and R2 is in parallel and then it is in series with R3

so parallel combination will be

r_1 = \frac{1*2.2}{2.2 +1} = 0.6875k ohm

also after its series with R3

r_2 = 1 + 0.6875 = 1.6875 k ohm

now it is in parallel with R5 on other side

r_3 = \frac{1.6875 * 2.2}{1.6875 + 2.2} = 0.95 k ohm

now current through the battery will be given as

i = \frac{3}{1 + 0.95} = 1.53 mA

now it is divide in r2 and R5

so current in R5 is given as

i_5 = \frac{r_2}{r_2 + R_5}*i

i_5 = \frac{1.6875}{2.2 + 1.6875} * 1.53

i_5 = 0.67 mA

now the total current in R5 will be given by super position which is

i = 0.67 + 0.18 = 0.85 mA

so there is 0.85 mA current through R5 resistance

8 0
3 years ago
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