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emmainna [20.7K]

3 years ago

6

What properties of radio waves makes them suitable for sending signals in mountainous regions?

1 answer:

marusya05 [52]3 years ago

3 0

one of the property of radio waves is that they have long wavelength. due to their long wavelength , they undergo diffraction around the mountains and thus the waves spread around the mountain. this makes them eligible to be used for long distance communication. they can coded and decoded also . that is why they are used in television and radio system

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Which process is the most scientifically plausible explanation for how the organisms released oxygen into Earth's atmosphere? Gr

Stels [109]

Answer:

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Explanation:

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The organism mainly responsible for this is known as cyanobacteria, or blue-green algae. These microbes conduct photosynthesis: using sunshine, water and carbon dioxide to produce carbohydrates and, oxygen.

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3 years ago

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3 years ago

Which contributions did Johannes Kepler make? Check all that apply. He revived Aristotle's model of the solar system. He solved

Arada [10]

The correct answers are: Options 2,4 and, 5

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3 0

3 years ago

Read 2 more answers

Un automovil parte del reposo y acelera uniformemente hasta alcanzar una rapidez de 0,255km/h en un tiempo de 3/4 Minutos determ

Elden [556K]

Answer:

a = 1.5*10^-3 m/s^2

x = 0.033m = 3.3cm

Explanation:

To calculate the acceleration and the distance traveled by the car you use the following formulas:

$v=v_o+at$ (1)

$x=v_ot+\frac{1}{2}at^2$ (2)

v: final velocity = 0,255 km/h

vo: initial velocity = 0 m/s

t: time = 3/4 min

a: acceleration = ?

x: distance

In order to use the equations (1) and (2) you first convert the units of the final velocity to m/s, and the time to seconds.

$v=0,255\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}\\\\v=0.07m/s\\\\t=\frac{3}{4}min*\frac{60s}{1min}=45s$

Next, you solve the equation (1) for the acceleration a:

$a=\frac{v}{t}=\frac{0.07m/s}{45s}=1.5*10^{-3}\frac{m}{s^2}$

With this value of a you can calculate the distance traveled by the car, by using the equation (2):

$x=\frac{1}{2}(1.5*10^{-3}m/s^2)(45s)^2=0.033m=3.3cm$

hence, the acceleration of the car is 1.5*10^-3 m/s^2 and the distance traveled in 3/4 min is 0.033m

5 0

3 years ago

Two spaceships are observed from earth to be approaching each other along a straight line. Ship A moves at 0.40c relative to the

nirvana33 [79]

Answer:

0.80 c

Explanation:

The computation of speed is shown below:-

Here, The speed of the captain ship A report for speed of the ship B which is

$S = \frac{S_A + S_B}{1 + \frac{(S_AS_B)}{c^2} }$

where

$S_A$ indicates the speed of the ship A

$S_B$ indicates the speed of the ship B

and

C indicates the velocity of life

now we will Substitute 0.40c for A and 0.60 for B in the equation which is

$S = \frac{0.40c + 0.60c}{1 + \frac{(0.40c)(0.60c)}{c^2} }$

after solving the above equation we will get

0.80 c

So, The correct answer is 0.80c

5 0

3 years ago