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maksim [4K]
3 years ago
15

Radio waves are readily diffracted around buildings, whereas light waves are negligibly diffracted around buildings. This is bec

ause radio waves
Physics
1 answer:
Tom [10]3 years ago
6 0

Answer:

Radio waves have longer wavelength

Explanation:

Radio wave is an electromagnetic frequency that has the ability to travel through long distance. They have frequencies shuttling been the range of 10^4 hz and a frequency of 10^12 hz

Light wave is also called visible light. This is because it is visible to the naked eye, despite it being in the electromagnetic spectrum. It's frequency is usually between 4*10^-7 hz and a frequency of 7*10^-7 hz.

As can be seen from both, the radio waves length are quite far stronger than that of the light waves.

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Suzanne bikes 3 km to school every day. it takes her 20 minutes because she always has to stop at a traffic light. what is her a
masha68 [24]
3 km in 20 min => 9 km in 60 min => 9km in 1h =>
 the correct answer is d: 9 km/h


3 0
3 years ago
A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 14 m/s. A passenger accidentally drops his camera fr
fgiga [73]

Answer:

<em>The balloon is 66.62 m high</em>

Explanation:

<u>Combined Motion </u>

The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is

\displaystyle y=y_o+v_o\ t-\frac{g\ t^2}{2}

The values are

\displaystyle y_o=15\ m

\displaystyle v_o=14\ m/s

We must find the values of t such that the height of the camera is 0 (when it hits the ground)

\displaystyle y=0

\displaystyle y_o+v_o\ t-\frac{g\ t^2}{2}=0

Multiplying by 2

\displaystyle 2y_o+2v_ot-gt^2=0

Clearing the coefficient of t^2

\displaystyle t^2-\frac{2\ V_o}{g}\ t-\frac{2\ y_o}{g}=0

Plugging in the given values, we reach to a second-degree equation

\displaystyle t^2-2.857t-3.061=0

The equation has two roots, but we only keep the positive root

\displaystyle \boxed {t=3.69\ s}

Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is

\displaystyle Y_r=15+V_r.t

\displaystyle Y_r=15+14\times3.69

\displaystyle \boxed{Y_r=66.62\ m}

3 0
3 years ago
A stockroom worker pushes a box with mass 11.2 kg on a horizontal surface with a constant speed of 3.30 m/s . The coefIficient o
Artemon [7]

Answer:

Explanation:

Mass =11.2kg

Constant velocity =3.3m/s

μk=0.25

Since the body is moving in constant velocity, then the acceleration is zero(0).

ΣF = Σ(ma)

The normal force acting on the body is upward and the weight is acting downward

Then ΣFy=0

Therefore, N=W

W=mg=11.2×9.8=109.76N

So, N=W=109.76N

Frictional force is given as

Fr=μkN

Fr=0.25×109.76

Fr=27.44N

Frictional force acting against the motion is 27.44N

Then the forward force moving the body forward

ΣF = Σ(ma)

Since a = 0

Then,

ΣF = 0

F-Fr=0

Then F=Fr

So the force moving the body forward is 27.44N

8 0
3 years ago
Analysis: (use a separate sheet of paper)
Zolol [24]

Speed, frequency and wavelength are interconnected to each other.

<h3>What happens to the wavelength if the frequency increases?</h3>

As a wavelength increases in size, its frequency and energy (E) decrease. As the frequency increases, the wavelength gets shorter. As the frequency decreases, the wavelength gets longer.

The wave speed is equal to the product of its frequency and wavelength, which shows the relationship between frequency and wavelength.

So we can conclude that speed, frequency and wavelength are interconnected to each other.

Learn more about frequency here: brainly.com/question/254161

5 0
2 years ago
The absolute magnitude of a star________.
serg [7]
<span>The absolute magnitude of a star is how bright it would appear to us
if it were located ten parsecs (about 32.6 light years) from us.  So it's
a way of treating all stars equally ... on a "level playing field" ... and it
describes each star's actual brightness. </span>
8 0
3 years ago
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