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s344n2d4d5 [400]
3 years ago
13

Name two things radio waves have in common with visible light

Physics
1 answer:
kondaur [170]3 years ago
8 0
Radio waves and visible light, as we perceive it, are both part of the EM spectrum. The only difference between them in terms of our perception is that we don't have receptors to 'see' radio waves. That's where radio receivers come into the picture*, to convert radio waves into sound impulses that we can hear.
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Listed following are the names and mirror diameters for six of the world’s greatest reflecting telescopes used to gather visible
ziro4ka [17]

Answer:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope

Explanation:

How much light a telescope can collect depends on its diameter, since in a bigger area more photons will be collected.    

Remember that in a circle the area is defined as:

A = \pi r^{2}  (1)

Where A is the area and r is its radius.

However, the radius can be determined by means of its diameter.

     

d = 2r

r = \frac{d}{2} (1)

Where d is its diameter.

An example of this is when a person is collecting raindrops with a bucket and with a cup. Since the bucket has a bigger area than the cup, it will collect more raindrops by unit of time. In this scenario the raindrops represent the photons.  

   

To determine the light collecting area of each telescope, equation 2 will be replaced in equation 1.

A = \pi (\frac{d}{2})^{2}  (3)

Case for Large binocular telescope:

A_{mirror1} = \pi (\frac{8.4m}{2})^{2}    

A_{mirror1} = 55.41m        

For the second mirror will be the same value

A = A_{mirror1}+A_{mirror2}  

A = 55.41m+55.41m

A= 110.82m

Case for Keck 1 telescope:

A = \pi (\frac{10m}{2})^{2}    

A = 78.53m  

Case for Hobby-Ebberly telescope:

A = \pi (\frac{9.2m}{2})^{2}    

A = 66.47m  

Case for Subaru telescope:

A = \pi (\frac{8.3m}{2})^{2}    

A = 54.10m  

Case for Gemini North telescope:

A = \pi (\frac{8m}{2})^{2}    

A = 50.26m  

Case for Magellan 2 telescope:

A = \pi (\frac{6.5m}{2})^{2}    

A = 33.18m  

Hence, they may be rank in the following way:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope.

<em>Key term:</em>

<em>Photons: particles that constitute light. </em>

3 0
3 years ago
The distance from one Crest to the next Crest is the blank
Luda [366]
The distance from one crest to the next crest in a set of waves is called the wavelength. The distance from the crest of one wave to the equilibrium point is called frequency.
Hope that helped! =)
8 0
3 years ago
Read 2 more answers
Newtons third law says that if Robert exerts a _______ of 1000 Newtons on an object, it will exert an equal and opposite _______
Free_Kalibri [48]

Answer: force, force

Explanation:

Newton’s third law states that there is an equal and opposite force

I took the test too

7 0
3 years ago
Can someone please explain number 8?
guajiro [1.7K]

Answer: Ax=(Vx-Vox)/(T)

Vx=Vox+Ax*T

Solving for Ax in terms of Vx, Vox, T

Vx-Vox=Ax*t

Ax=(Vx-Vox)/(T)

This is saying the acceleration in the x-direction can be found by taking the difference between the finial and initial Velocity in x-direction and dividing it by the Total Time.

Any questions please feel free to ask. Thanks

8 0
3 years ago
A bowling ball that has a radius of 11.0 cm and a mass of 5.00 kg rolls without slipping on a level lane at 2.80 rad/s.
NemiM [27]

Answer:

\dfrac{K_t}{K_r}=\dfrac{5}{2}

Explanation:

Given that,

Mass of the bowling ball, m = 5 kg

Radius of the ball, r = 11 cm = 0.11 m

Angular velocity with which the ball rolls, \omega=2.8\ rad/s

To find,

The ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball.

Solution,

The translational kinetic energy of the ball is :

K_t=\dfrac{1}{2}mv^2

K_t=\dfrac{1}{2}m(r\omega)^2

K_t=\dfrac{1}{2}\times 5\times (0.11\times 2.8)^2

The rotational kinetic energy of the ball is :

K_r=\dfrac{1}{2}I \omega^2

K_r=\dfrac{1}{2}\times \dfrac{2}{5}mr^2\times \omega^2

K_r=\dfrac{1}{2}\times \dfrac{2}{5}\times 5\times (0.11)^2\times (2.8)^2

Ratio of translational to the rotational kinetic energy as :

\dfrac{K_t}{K_r}=\dfrac{5}{2}

So, the ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is 5:2

4 0
3 years ago
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