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soldier1979 [14.2K]
3 years ago
9

Which part of the electromagnetic spectrum generally gives us our best views of stars forming in dusty clouds?

Physics
1 answer:
Inessa05 [86]3 years ago
3 0
<span>We can see stars forming in dusty clouds because of the infrared portion of the electromagnetic spectrum. Infrared waves pass through dust in outer space more easily than visible light. Interstellar dust absorbs visible light, a phenomena known as dust extinction.</span>
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Can someone please help me?? i’ll give brainilist
Ksivusya [100]
Probably 90 j but im not sure I haven’t done any work like this in a while
7 0
3 years ago
The 1.18-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in
sattari [20]

Answer:

 k = 11,564 N / m,   w = 6.06 rad / s

Explanation:

In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;

 let's apply the equilibrium condition at this point

                 

Axis y

          W_{y} - Fr = 0

          Fr = k y

let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal

             sin 46 = W_{y} / W

             W_{y} = W sin 46

     

 we substitute

           mg sin 46 = k y

           k = mg / y sin 46

If the length of the bar is L

          sin 46 = y / L

           y = L sin46

 

we substitute

           k = mg / L sin 46 sin 46

           k = mg / L

for an explicit calculation the length of the bar must be known, for example L = 1 m

           k = 1.18 9.8 / 1

           k = 11,564 N / m

With this value we look for the angular velocity for the point tea = 30º

let's use the conservation of mechanical energy

starting point, higher

          Em₀ = U = mgy

end point. Point at 30º

         Em_{f} = K -Ke = ½ I w² - ½ k y²

          em₀ = Em_{f}

          mgy = ½ I w² - ½ k y²

          w = √ (mgy + ½ ky²) 2 / I

the height by 30º

           sin 30 = y / L

           y = L sin 30

           y = 0.5 m

the moment of inertia of a bar that rotates at one end is

          I = ⅓ mL 2

          I = ½ 1.18 12

          I = 0.3933 kg m²

let's calculate

          w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)

          w = 6.06 rad / s

7 0
3 years ago
What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? a plane accelerates from
ElenaW [278]
When is at the end of the runway the velocity of the plane is given by the equation vf^{2}=0+2*a*s    where s=1800 m is the runway length. Thus
vf^{2}=2*5*1800=18000 (m/s)^{2}      
vf =134.164 (m/s)  

At half runway the velocity of the plane is
v^{2}=2*5* \frac{1800}{2}=9000 ( \frac{m}{s} )^{2}&#10; 
v= \sqrt{9000}=94.87 ( \frac{m}{s})

Therefore at midpoint of runway the percentage of takeoff velocity is
‰P= \frac{v}{vf}=  \frac{94.87}{134.164}=0.707
6 0
3 years ago
A soap bubble appears red (λ = 633nm) at the point on its front surface nearest to the viewer. Assuming n = 1.35, what is the sm
alexira [117]

Answer:

The smallest film thickness is 117 nm.

Explanation:

Light interference on thin films can be constructive or destructive. Constructive interference is dependent on the film thickness and the refractive index of the medium.

For the first interference (surface nearest to viewer), the minimum thickness can be expressed as:

2t_{min} = \frac{wavelenth}{2n}

where n is the refractive index of the bubble film.

Therefore,

2t_{min} = \frac{633x10^{-9} }{(2)(1.35)}

2t_{min} =2.344x10^{-7}

∴ t_{min} =\frac{2.344x10^{-7} }{2}

t_{min} = 1.17x10^{-7} m = 117 nm.

7 0
3 years ago
Answer??????????????
solong [7]
The number of protons!!
3 0
3 years ago
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