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nadezda [96]
3 years ago
12

The frequency of a given region of the electromagnetic spectrum ranges from 3 × 1016 − 3 × 1019 hertz. Which type of wave is fou

nd in this region?
A. x-rays
B. ultraviolet rays
C. infrared waves
D. radio waves
Physics
2 answers:
Bond [772]3 years ago
5 0
<span>Wavelength = (speed) / (frequency)
Speed of EM radiation = 3 x 10⁸ m/s

Frequency (3 × 10¹⁹ Hz)  ===>  wavelength  =  0.01 nanometer

Frequency (3 × 10¹⁶ Hz)  ===>  wavelength  =  10 nanometers</span>

This is the region of X-rays.
myrzilka [38]3 years ago
4 0

The correct answer to the question is : A) X -rays.

EXPLANATION:

As per the question, the frequency of the given radiation is 3\times 10^{16} -\ 3\times 10^{19}\ Hz..

If we arrange all the electromagnetic radiations in the ascending order of their frequency, then the given frequency is less than the frequency of gamma wave and greater than U.V ray.

The given frequency corresponds to the frequency of X-rays in the electromagnetic spectrum.

Hence, the correct answer of the question is X-rays.

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What do nebulae have the most in common with?
alexdok [17]
Most commonly the answer is A, the sun. A nebula is a nursery for stars. The Sun itself was created in a nebula. Also,you do not need college physics to understand this question


3 0
3 years ago
In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.
Rzqust [24]

1) 0.0011 rad/s

2) 7667 m/s

Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

\omega=\frac{\theta}{t}

where

\theta is the angular displacement of the object

t is the time elapsed

\omega is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

\theta=2\pi rad

And the time taken is

t=95 min \cdot 60 =5700 s

Therefore, the angular velocity of the telescope is

\omega=\frac{2\pi}{5700}=0.0011 rad/s

2)

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

v=\omega r

where

v is the linear velocity

\omega is the angular velocity

r is the radius of the circular orbit

In this problem:

\omega=0.0011 rad/s is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m

Therefore, the linear velocity of the telescope is:

v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s

4 0
3 years ago
Hat processes lead to glacial erosion ?
azamat
Yes......................
8 0
3 years ago
Lexington walked to her friends house.lexie walked 2miles west then 5 miles south then 3 miles east then 4 miles north then 2 mi
vampirchik [111]
The answer is 17 for the miles
4 0
3 years ago
can somebody help me answer this question? A car went from 110 m/s to 80 m/s in 20 seconds. What was the acceleration of the car
gogolik [260]

Answer:

-1.5m/s²

Explanation:

Acceleration can be thought of as [Change in Velocity]/[Change in time]. To find these changes, you simply subtract the initial quantity from the final quantity.

So for this question you have:

  • V_i = 110m/s
  • V_f = 80m/s
  • t_i = 0s
  • t_f = 20s

which means that the acceleration = (80-110)/(20-0)[m/s²] = (-30/20)m/s² = -1.5m/s²

4 0
3 years ago
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