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Firlakuza [10]
3 years ago
15

If we aim a radio telescope at a distant spiral arm of the Milky Way Galaxy, we will probably observe a 21-cm line. If we point

a large optical telescope at this same region, we will probably not be able to detect the neutral hydrogen that gives rise to the 21-cm radio signal. Why not
Physics
1 answer:
Oxana [17]3 years ago
5 0

Answer:

Because of the interstellar dust and interior location of the solar system.

Explanation:

We will probably not be able to detect the neutral hydrogen that gives rise to the 21-cm radio signal if we point a large optical telescope to the region because, the interstellar dust obscures the location of the spiral arm of the Milky way galaxy and this makes neutral hydrogen that gives rise to the 21-cm radio signal difficult to detect.

Also, the interior location of the solar system also makes the neutral hydrogen that gives rise to the 21-cm radio signal difficult to detect.

<u>So, the interstellar dust and the interior location of the solar system makes it difficult to detect the neutral hydrogen that gives rise to the 21-cm radio signal with a large optical telescope.</u>

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A sample of copper has a volume of 23.4 cm3 if the density of copper is 8.9 gcm3 what is the coppers mass?
murzikaleks [220]
The answer is:  " 208 g " .
_____________________________________________
Explanation:
__________________________________________
The formula/ equation for density is:
__________________________________________
D = m / V  ;  That is,  "mass divided by volume" ;
 
Density is expressed as:
__________________________________________    
                   "mass per unit volume";  in which the "mass" is expressed in units of "g" ("grams") ;  and the "unit volume" is expressed in units of:
    "cm³ " or "mL"; 
_____________________________________________
           {Note the exact equivalent:  1 cm³ = 1 mL }.
____________________________________________
         →  The formula is:  " D = m / V "  ; 
___________________________________________
   in which:

     "D" refers to the "density" (see above), which is: "8.9 g/cm³ " (given); 

     "m" refers to the "mass" , in units of "g" (grams), which is unknown; and we want to find this value;
                 
     "V" refers to the "volume", in units of "cm³ " ;
               which is:  "23.4 cm³ " (given);
_________________________________________________
                 We want to find the mass, "m" ; so we take the original equation/formula for the density:
_________________________________________________ 
              D  =  m / V ; 
_________________________________________________________
             And we rearrange; to isolate "m" (mass) on ONE side of the    equation; and then we plug in our known/given values;
 to solve for "m" (mass);  in units of "g" (grams) ;
___________________________________________________
    Multiply each side of the equation by "V" ; 
____________________________________________________
             V * { D  =  m / V } ;  to get:
____________________________________________________
      V * D = m ;   ↔   m = V * D ;
___________________________________________________
           Now, we plug in the given values for "V" (volume) and "D" (density) ;     to solve for the mass, "m" ;
______________________________________________________
           m  =  V * D ;
 
           m  =  (23.4 cm³) * (8.9 g / 1 cm³)  = (23.4 * 8.9) g = 208.26 g ;
  
 →  Round to "208 g" (3 significant figures);  
____________________________________
The answer is:  " 208 g " .
_____________________________________________________
7 0
3 years ago
The nuclei of large atoms, such as uranium, with 9292 protons, can be modeled as spherically symmetric spheres of charge. The ra
Scrat [10]

Answer:

Part 1 E = 2.42 * 10^{21}  N/C

Part 2 E = 1.3 * 10^{13}  N/C

Part 3 E = 0

Explanation:

Given

Number of protons = 92

Radius of nucleus r_n = 7.4 * 10^{-15} m

Distance of the electrons r_1 = 1.0 * 10^ {-10} m

Part 1

Electric field produced by  just outside its surface

E  = \frac{q}{4\pi*E_0* r_n^2 } \\E  = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(7.4* 10^{-15})^2} \\E = 2.42 * 10^{21}  N/C

Part 2

Electric field produced by  just outside its surface

E  = \frac{q}{4\pi*E_0* r_n^2 } \\E  = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(1* 10^{-15})^2} \\E = 1.3 * 10^{13}  N/C

Part 3

The net electric field inside a uniform shell of negative charge is zero because the electric flux lines cancel out each other

hence, the solution is

Part 1 E = 2.42 * 10^{21}  N/C

Part 2 E = 1.3 * 10^{13}  N/C

Part 3 E = 0

7 0
3 years ago
Continuous sinusoidal perturbation Assume that the string is at rest and perfectly horizontal again, and we will restart the clo
Elena-2011 [213]

a) 3.14 \cdot 10^{-4} s

b) See plot attached

c) 10.0 m

d) 0.500 cm

Explanation:

a)

The position of the tip of the lever at time t is described by the equation:

y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t] (1)

The generic equation that describes a wave is

y(t)=A sin (\frac{2\pi}{T} t) (2)

where

A is the amplitude of the wave

T is the period of the wave

t is the time

By comparing (1) and (2), we see that for the wave in this problem we have

\frac{2\pi}{T}=2.00\cdot 10^4 s^{-1}

Therefore, the period is

T=\frac{2\pi}{2.00\cdot 10^4}=3.14 \cdot 10^{-4} s

b)

The sketch of the profile of the wave until t = 4T is shown in attachment.

A wave is described by a sinusoidal function: in this problem, the wave is described by a sine, therefore at t = 0 the displacement is zero, y = 0.

The wave than periodically repeats itself every period. In this sketch, we draw the wave over 4 periods, so until t = 4T.

The maximum displacement of the wave is given by the value of y when sin(...)=1, and from eq(1), we see that this is equal to

y = 0.500 cm

So, this is the maximum displacement represented in the sketch.

c)

When standing waves are produced in a string, the ends of the string act as they are nodes (points with zero displacement): therefore, the wavelength of a wave in a string is equal to twice the length of the string itself:

\lambda=2L

where

\lambda is the wavelength of the wave

L is the length of the string

In this problem,

L = 5.00 m is the length of the string

Therefore, the wavelength is

\lambda =2(5.00)=10.0 m

d)

The amplitude of a wave is the magnitude of the maximum displacement of the wave, measured relative to the equilibrium position.

In this problem, we can easily infer the amplitude of this wave by looking at eq.(1).

y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t]

And by comparing it with the general equation of a wave:

y(t)=A sin (\frac{2\pi}{T} t)

In fact, the maximum displacement occurs when the sine part is equal to 1, so when

sin(\frac{2\pi}{T}t)=1

which means that

y(t)=A

And therefore in this case,

y=0.500 cm

So, this is the displacement.

6 0
3 years ago
The radius of Earth is . The average Earth-Sun distance is . How many Earths would fit between Earth and the Sun if they are sep
Verdich [7]

Answer:

The radius of the earth is 6,371 km.

The average Earth-Sun distance is 152.09 million km

How many Earths would fit between Earth and the Sun if they are separated by their average distance? Approximately 11,936 Earths.

I didn't really understand the last part, but if you don't get a better answer please mark me as brainliest.

6 0
3 years ago
Two black holes (the remains of exploded stars), separated by a distance of
jolli1 [7]

The largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

The given parameters;

  • <em>distance between the two black holes, r = 10 AU = 1.5 x 10¹² m</em>
  • <em>gravitational force between the two black holes, F = 6.9 x 10²⁵ N.</em>
  • <em>combined mass of the two black holes = 5.20 x 10³⁰ kg</em>

The product of the two masses is calculated from Newton's law of universal gravitational as follows;

F = \frac{Gm_1m_2}{r^2} \\\\m_1m_2 = \frac{Fr^2}{G} \\\\m_1m_2 = \frac{(6.9\times 10^{25}) \times (1.5\times 10^{12})^2}{6.67\times 10^{-11}} \\\\m_1m_2 = 2.328 \times 10^{60} \ kg^2

The sum of the two masses is given as;

m₁ + m₂ = 5.2 x 10³⁰ kg

m₂ = 5.2 x 10³⁰ kg - m₁

The first mass is calculated as follows;

m₁(5.2 x 10³⁰ - m₁) = 2.328 x 10⁶⁰

5.2 x 10³⁰m₁ - m₁² = 2.328 x 10⁶⁰

m₁² - 5.2 x 10³⁰m₁  + 2.328 x 10⁶⁰ = 0

<em>solve the quadratic equation using formula method</em>;

a = 1, b =-  5.2 x 10³⁰, c = 2.328 x 10⁶⁰

m_1 = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\m_1 = \frac{-(-5.2\times 10^{20})  \ \ +/- \ \ \sqrt{(-5.2\times 10^{20})^2 - 4(1\times 2.328\times 10^{60})} }{2(1)} \\\\m_1 = 4.7 \times 10^{30} \ kg \ \ or \ \ 4.9 \times 10^{29} \ kg

The second mass is calculated as follows;

m₂ = 5.2 x 10³⁰ kg - m₁

m₂ = 5.2 x 10³⁰ kg  -  4.7 x 10³⁰ kg

m₂ = 5 x 10²⁹ kg

or

m₂ = 5.2 x 10³⁰ kg  -  4.9 x 10²⁹ kg

m₂ = 4.7 x 10³⁰ kg

Thus, the largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

Learn more here:brainly.com/question/9373839

3 0
2 years ago
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