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alex41 [277]
3 years ago
7

A quasar is a distant celestial body in space. Investigators use a special telescope and determine that a certain quasar was giv

ing off waves with a frequency of 1.41 X 109 Hz. An electromagnetic spectrum is shown above. The investigators were most likely using a telescope that detects which type of wave?
Physics
1 answer:
iris [78.8K]3 years ago
6 0

Answer:

equipment for the measurement of microwave L bands with a range between 1 GHz and 2 GHz.

Explanation:

The electromagnetic spectrum can be calculated with the relationship between the speed of light, its wavelength and its frequency.

         c = λ f

For reasons of analysis and equipment used, it is artificially divided into ranges, with poorly defined limits and in some cases with overlaps between some.

For the case of analysis, f = 1.41 10⁹ Hz, we have the range called

* Microwave for f> 3 108 Hz to approximately f <3 1011 Hz

For the lower part of the frequency 3 10⁸ <f <3 10⁹ Hz we have UHF television channels and cell phones and military communications.

As the frequency observed by the researchers is in the UHF range, it is possible that they are using microwave equipment for communications, specifically equipment for the measurement of microwave L bands with a range between 1 GHz and 2 GHz.

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The power lines are at a high potential relative to the ground, so there is an electric field between the power lines and the gr
Amanda [17]

Answer:

The tube should be held vertically and perpendicular to the ground.

Explanation:

Answer: The tube should be held vertically and perpendicular to the ground. The reason is as follows:

Reasoning:

The power lines are parallel to the ground hence, their electric field will be perpendicular to the ground and equipotential surface will be cylindrical.

Hence, if you will put fluorescent tube parallel to the ground then both the ends of the tube will lie on the same equipotential surface and the potential difference will be zero.

So, to maximize the potential the ends of the tube must be on different equipotential surfaces. The surface which is near to the power line has high potential value and the surface which is farther from the line has lower potential value.

hence, to maximize the potential difference, the tube must be placed perpendicular to the ground.

6 0
2 years ago
A box weighing 12,000 N is parked on a 36° slope. What will be the component of the weight parallel to the plane that balances f
sweet-ann [11.9K]

Given :

A box weighing 12,000 N is parked on a 36° slope.

To Find :

What will be the component of the weight parallel to the plane that balances friction.

Solution :

The component of that will be parallel to the plane to balance friction is :

F \ cos ( 90 - 36)^o\\\\F\ ( sin\ 36^o)

Therefore, component of force to balance friction is F sin 36° .

Hence, this is the required solution.

5 0
2 years ago
Explain why average velocity in one dimension can be positive or negative.
Softa [21]
Imagine an object is moving in one dimension on a number line, and for this we'll say that the numbers on the line are a metre apart. If the object moves from 2 m to 7 m, the change in position is 7-2=+5 metres. But if the object moves back from 7 m to 2 m, the change in position is 2-7=-5 metres. since velocity =  \frac{change in position}{time}, and time is always positive, velocity will be positive in one direction and negative in the other direction.
3 0
3 years ago
Read 2 more answers
At a hot air balloon race, a person on the ground shots a ball of confetti from a cannon to start the race. One of the balloons
Sati [7]

Answer:

a) The balloon and the ball will meet after 1.43 s.

b) The ball will reach the balloon at 15.7 m above the ground.

Explanation:

The height of the confetti ball is given by the following equation:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the ball at time t

y0 = initial height

v0 = initial velocity

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

The height of the ball is given by this equation:

y = y0 + v · t

Where v is the constant velocity.

When the ball and the ballon meet, both heights are equal. Let´s consider the ground as the origin of the frame of reference so that y0 = 0:

y balloon = y ball

y0 + v · t = y0 + v0 · t + 1/2 · g · t²                  (y0 = 0)

11 m/s · t = 18 m/s · t -1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 18 m/s · t - 11 m/s · t

0 = -4.9 m/s² · t²  + 7 m/s · t

0 = t( -4.9 m/s² · t  + 7 m/s)

t = 0 and

0 = -4.9 m/s² · t  + 7 m/s

-7 m/s / - 4.9 m/s² = t

t = 1.43 s

They will meet after 1.43 s

b) Now let´s calculate the height of the balloon after 1.43 s

y = v · t

y = 11 m/s · 1.43 s = 15.7 m

The ball will reach the balloon at 15.7 m above the ground.

7 0
3 years ago
PLEASE HELP FOR PHYSICS!
Stolb23 [73]

Hi there!

Recall Newton's Law of Universal Gravitation:

\large\boxed{F_g = G\frac{m_1m_2}{r^2}}

Where:

Fg = Force of gravity (N)

G = Gravitational Constant

m1, m2 = masses of objects (kg)

r = distance between objects (m)

Plug in the given values stated in the problem:

F_g = (6.673*10^{-11})\frac{50 * 50}{25^2} = \boxed{2.669 * 10^{-10} N}

8 0
2 years ago
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