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masha68 [24]
3 years ago
7

Mike recently purchased an optical telescope. Identify the part of the electromagnetic spectrum that is closest to the frequency

that Mike can observe with the help of his new tool.

Physics
2 answers:
Zepler [3.9K]3 years ago
8 0
The elctromagnetic spectrum ranges from the radiowaves to the gamma rays. The whole spectrum is shown in the attached picture. But the optical telescope can only see the visible region. So, it only covers from the 400 nm to 700 nm frequency. It follows the ROYGBIV colors, where red has the highest frequency and violet has the lowest frequency.

zalisa [80]3 years ago
7 0

Answer:

Ultraviolet and Infrared

Explanation:

Electromagnetic spectrum shows all type of EM waves basis their frequency and wavelength. It contains EM waves from radio waves having the smallest frequency to gamma rays having maximum frequency. Through an optical telescope Mike can observe only the visible part of the spectrum which is just 7% of the complete spectrum.

The closest to the frequency range of visible light lies Ultraviolet (UV) and Infrared rays. UV rays have higher frequency than visible light and IR rays have lesser frequency.

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The amount of light that enters the pupil is controlled by the:<br> retina.<br> lens.<br> inis.
nignag [31]

Answer: The amount of light that enters the pupil is controlled by the Iris

Explanation:

5 0
3 years ago
Riders in a carnival ride stand with their backs against the wall of a circular room of diameter
Veseljchak [2.6K]

Answer:

option C

Explanation:

given,

diameter of circular room = 8 m

rotational velocity of the rider = 45 rev/min

                  = 45 \times \dfrac{2\pi}{60}

                  =4.712 rad/s

here in this case normal force is equal to centripetal force

N = m r ω²

N = m x 4 x 4.712²

N = 88.83m

frictional force = μ N

    = 88.83m x μ

now, for the body to not to slide

gravity force is equal to frictional force

m g = 88.83 m x μ

g = 88.83 x μ

9.8 = 88.83 x μ

 μ = 0.11

hence, the correct answer  is option C

6 0
4 years ago
A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur
muminat

Answer:

a) E = 0

b) E =  \dfrac{k_e \cdot q}{ r^2 }

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

From which we have;

E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

E = 0/A = 0

E = 0

b) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

E \cdot  A  = \dfrac{+q }{\varepsilon _{0}}

E  = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}

By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

Therefore, we get;

E \cdot (4 \cdot \pi \cdot r^2) =  \dfrac{q}{\varepsilon _{0}}

The electrical field outside the spherical shell

E =  \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }=  \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }

k_e=  \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }

Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

5 0
3 years ago
Determine the gain in the potential energy when a 8.0 kg box is raised 17.2 m.
Marysya12 [62]

Answer:

<h2>The answer is 1376 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 8 × 10 × 17.2

We have the final answer as

<h3>1376 J</h3>

Hope this helps you

6 0
3 years ago
Read 2 more answers
Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.0 kg gibbon
LiRa [457]

Answer:

241.8 N.

Explanation:

The force on branch provides a reaction to the ape's weight force plus the centripetal force needed to keep the gibbon in a circular motion of radius 0.60 m.

Centripetal force = mv^2/r

F = mg + mv²/r

F = m(g + v²/r)

where,

m = mass

= 9 kg

g = acceleration due to gravity

= 9.8 m/s²

v = 3.2 m/s

r = 0.60 m

F = 9 * (9.8 + 3.2²/0.60)

= 241.8 N.

3 0
3 years ago
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