answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
masha68 [24]
2 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]2 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]2 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.

You might be interested in
An infinite conducting cylindrical shell of outer radius r1 = 0.10 m and inner radius r2 = 0.08 m initially carries a surface ch
gayaneshka [121]

Answer:

Explanation:

Solution is in the picture attached

8 0
2 years ago
As an object in motion becomes heavier, its kinetic energy _____. A. increases exponentially B. decreases exponentially C. incre
diamong [38]

Answer:

USE SOCRACTIC IT WOULD REALLY HELP

4 0
2 years ago
Read 2 more answers
Table 2.4 shows how the dispacement of a runner changed during a sprint race. Draw a dispacement-time graph to show this data, a
GalinKa [24]
4. Table 2.4 shows how the displacement of a runner changed
during a sprint race. Draw a displacement–time graph to show
this data, and use it to deduce the runner’s speed in the middle
of the race.
Table 2.4 Data for a sprinter during a race
Displacement
(m)
0 4 10 20 50 80 105
Time (s) 1 2 3 6 9 12
8 0
1 year ago
An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t
gtnhenbr [62]

1. tan \theta = \frac{v^2}{Rg}

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

N cos \theta = mg (1)

N sin \theta = m \frac{v^2}{R} (2)

where in the second equation, the term m\frac{v^2}{R} represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the  following expression:

tan \theta = \frac{v^2}{Rg}

2. F=\frac{m}{R}(w^2-v^2)

In this second situation, the cars moves around the track at a speed

w>v

This means that the centripetal force term

m\frac{v^2}{R}

is now larger than before, and therefore, the horizontal component of the normal reaction, N sin \theta, is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

N sin \theta + F = m\frac{w^2}{R}

And re-arranging for F,

F=m\frac{w^2}{R}-N sin \theta (3)

But from eq.(2) in the previous part we know that

N sin \theta = m \frac{v^2}{R}

So, susbtituting into eq.(3),

F=m\frac{w^2}{R}-m\frac{v^2}{R}=\frac{m}{R}(w^2-v^2)

4 0
2 years ago
You travel in a circle, whose circumference is 8 kilometers, at an average speed of 8 kilometers/hour. If you stop at the same p
Schach [20]
Velocity = (displacement) / (time)

Displacement = straight-line distance between start-point and end-point

If you stop at the same point you started from, then
your displacement for the trip is zero, and your average
velocity is also zero.

5 0
2 years ago
Read 2 more answers
Other questions:
  • The number that is used to show the value of one currency compared to another is called the __________. A. trade rate B. currenc
    6·1 answer
  • What is the value of the composite constant (gmer2e), to be multiplied by the mass of the object mo in the equation above? expre
    9·2 answers
  • It takes 87 j of work to stretch an ideal spring, initially 1.4 m from equilibrium, to a position 2.9 m from equilibrium. what i
    13·1 answer
  • A small cork with an excess charge of +6.0µC is placed 0.12 m from another cork, which carries a charge of -4.3µC.
    11·1 answer
  • what is a possible unit for the product VI, where V is the potential difference across a resistor and I is the current through t
    14·1 answer
  • A cartridge electrical heater is shaped as a cylinder of length L = 200 mm and outer diameter D = 20 mm. Under normal operating
    10·1 answer
  • In an experiment, one of the forces exerted on a proton is F⃗ =−αx2i^, where α=12N/m2. What is the potential-energy function for
    12·1 answer
  • A series circuit contains an 80-μF capacitor, a 0.020-H inductor, and a switch. The resistance of the circuit is negligible. Ini
    14·1 answer
  • You are flying a hang glider at 14 mph in the northeast direction (45°). The wind is blowing at 4 mph from due north.
    11·1 answer
  • The Si unit of potential difference is a) volt b) JA⁻¹s⁻¹ c)JC⁻¹ d) All the above
    9·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!