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2 years ago

What are the wavelengths of electromagnetic wave in free space that have the following frequencies?.

Physics

2 answers:

irina [24]2 years ago

7 0

The wavelengths of the light are 4.3 * 10^-12 m and 0.2 m respectively.

<h3>What is wavelength?</h3>

The term wavelength has to do with the horizontal distance that is covered by a wave. We know that a long wavelength implies that the wave is able to travel a long distance from one point to another.

Given that;

c = λf

c = speed of light

λ = wavelength of ight

f = frequency of light

Thus;

λ = 3 * 10^8/ 7.00 x 10^19

λ = 4.3 * 10^-12 m

λ = 3 * 10^8/1.50 x 10^9

λ = 2 * 10^-1 or 0.2 m

Learn more about wavelength:brainly.com/question/13533093

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Missing parts:

What are the wavelengths of electromagnetic wave in free space that have the following frequencies? (a) 7.00 x 10^19 Hz______ pm (b) 1.50 x 10^9 Hz__________ cm

natulia [17]2 years ago

4 0

~400 nm to ~700 nm are the wavelengths of electromagnetic wave in free space.

What are the wavelengths of electromagnetic wave in free space?

Electromagnetic waves are categorized on the basis of their frequency f, which is equivalent to wavelength λ = c/f. Visible light has a wavelength that ranges from ~400 nm to ~700 nm. Violet light has a wavelength of ~400 nm, and having a frequency of ~7.5*1014 Hz.

So we can conclude that: ~400 nm to ~700 nm are the wavelengths of electromagnetic wave in free space.

Learn more about electromagnetic wave here :brainly.com/question/25847009

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Definition of continental polar

kipiarov [429]

Answer:

Continental polar (cP) or continental arctic (cA) air masses are cold, dry, and stable. These air masses originate over northern Canada and Alaska as a result of radiational cooling. Maritime polar (mP) air masses are cool, moist, and unstable.

Explanation:

8 0

3 years ago

Read 2 more answers

A child uses a rubber band to launch a bottle cap at an angle of 37.0° above the horizontal. The cap travels a horizontal distan

zavuch27 [327]

Answer:

Initial velocity will be 1.356 m/sec

Explanation:

Let the initial speed = u

Angle at which rubber band is launched = 37°

Horizontal component of initial velocity $u_x=ucos\Theta =ucos37^{\circ}=0.7986u$

Time is given as t = 1.20 sec

Distance in horizontal direction = 1.30 m

We know that distance = speed × time

So time $t=\frac{distance}{speed}$

$1.20=\frac{1.3}{0.7986u}$

$u=1.356m/sec$

So initial velocity will be 1.356 m/sec

3 0

3 years ago

A car has a mass of 550 kg. What is the weight of the car?

lozanna [386]

1212.54 is what 550 kgs is equal to the weight of the car

3 0

3 years ago

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Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject materia

kramer

Answer:

The height reached by the material on Earth is 91 km.

Explanation:

Given that,

Mass $M_{Io}=8.93\times10^{22}\ kg$

Radius = 1821 km

Height $h_{Io}=500\ km$

Suppose we need to find that how high would this material go on earth if it were ejected with the same speed as on Io?

We need to calculate the acceleration due to gravity on Io

Using formula of gravity

$g =\dfrac{GM_{Io}}{(R_{Io})^2}$

Put the value into the formula

$g=\dfrac{6.67\times10^{-11}\times8.93\times10^{22}}{(1821\times10^{3})^2}$

$g=1.79\ m/s^2$

Let v be the speed at which the material is ejected.

We need to calculate the height

Using the formula of height

$H=\dfrac{v^2}{2g}$

Using ratio of height of earth and height of Io

$\dfrac{H_{e}}{H_{Io}}=\dfrac{\dfrac{v^2}{2g_{e}}}{\dfrac{v^2}{2g_{Io}}}$

$\dfrac{H_{e}}{H_{Io}}=\dfrac{g_{Io}}{g_{e}}$

Put the value into the formula

$\dfrac{H_{e}}{H_{Io}}=\dfrac{1.79}{9.8}$

$\dfrac{H_{e}}{H_{Io}}=0.182$

$H_{e}=0.182\times H_{Io}$

$H_{e}=0.182\times500$

$H_{e}=91\ km$

Hence, The height reached by the material on Earth is 91 km.

3 0

3 years ago

Alcohol consumption slows people's reaction times.In a controlled government test,it takes a certain driver 0.320 s to hit the b

svetoff [14.1K]

Answer:

When the person is drunk the car travels 20.1 meters more than when the person is sober. The distance traveled by car is:

Person drunk: 28.1 m

Person sober: 8 m

Explanation:

The distance traveled can be found using the following equation:

$v = \frac{x}{t} \rightarrow x = v*t$

Where:

v: is the speed = 90 km/h

t: is the time

When the person is sober, the time that takes to hit the brakes is 0.320 s, so we have:

$x = v*t = 90 \frac{km}{h}*\frac{1 h}{3200 s}*0.320 s = 8 \cdot 10^{-3} km = 8 m$

And when the person is drunk, the time is 1.00 s, hence the distance is:

$x = v*t = 90 \frac{km}{h}*\frac{1 h}{3200 s}*1 s = 0.0281 km = 28.1 m$

The distance traveled by the car when the person is drunk compared when the person is sober is:

$d = 28.1 m - 8 m = 20.1 m$

Therefore, when the person is drunk the car travels 20.1 meters more than when the person is sober.

I hope it helps you!

6 0

3 years ago