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3 years ago

How do very long radio waves reach stations beyond the curvature of the Earth?

2 answers:

8 0

They actually bounce off a charged layer of atmosphere to reach stations beyond the curvature of the Earth. The correct option among all the options that are given in the question is the last option or option "d". The radio waves bounce off the ions present in the ionosphere of earths atmosphere and reach stations beyond the curvature of Earth.

Drupady [299]3 years ago

5 0

The correct answer for the question that is being presented above is this one: "d. They bounce off a charged layer of the atmosphere." Very long radio waves reach stations beyond the curvature of the Earth. They bounce off a charged layer of the atmosphere

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Question in picture, just want to confirm answer. Is the answer B?

guapka [62]

Answer:

Explanation:

yep the answer is B

6 0

2 years ago

The reflective quality of a surface is known as its

Brut [27]

<h2>Answer: Albedo </h2>

The albedo is an amount that expresses the percentage of radiation a surface reflects with respect to the incident radiation.

In other words:

This amount allows us to know the level of radiation that reflects a surface compared to the total radiation it receives.

According to this, light surfaces such as snow covered ground or white sand will have a higher albedo than dark surfaces such as carbon covered ground. It is also important to note, the albedo will be higher on glossy surfaces than on matte surfaces.

It should be noted that the albedo of the Earth is on average about $37\%$ , which means that part of the radiation received by the Sun is absorbed and another part reflected back to space.

4 0

3 years ago

What do you think we can learn about Earth's history by looking at the Grand Canyon?

viktelen [127]

We can look at all the ages of the earth since it’s a big crack is reveals many layers of the earth and we can know about chemicals and metals that were in earth and diffrent times

5 0

3 years ago

A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,

kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

$\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}$

so that the x-component of $\vec v_{\rm ave}$ is

$\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}$

and its y-component is

$\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}$

Solve for $x_2$ and $y_2$ , which are the x- and y-components of the copter's position vector after t = 1.60 s.

$x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}$

$y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}$

Note that I'm reading the given details as

$x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}$

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

8 0

3 years ago

A rabbit is hopping along at an approximately constant speed of 3.9 m/s. The rabbit passes a crouched cat ready to chase the rab

Firlakuza [10]

Answer:

t = 7,8 s

Explanation:

From the instant, the rabbit passes the cat. The cat star running acceleration of 0,5 m/s² .

When the cat arrives at the speed of 3,9 m/s the cat catches the rabbit

Then for the cat arrives at 3,9 m/s nedds

v = vo + a*t vo = 0 then v = a*t

3,9 ( m/s) = 0,5 ( m/s² ) * t

t = 7,8 s

v = 3,9 m/s =

4 0

2 years ago