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3 years ago

A beam of light has a wavelength of 4.5 x10^-7 meter in a vacuum. the frequency of this light is

1 answer:

3 0

The basic relationship between frequency and wavelength for light (which is an electromagnetic wave) is
$c= f \lambda$
where c is the speed of light, f the frequency and $\lambda$ the wavelength of the wave.
Using $\lambda=4.5 \cdot 10^{-7} m$ and $c=3 \cdot 10^8 m/s$ , we can find the value of the frequency:
$f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{4.5 \cdot 10^{-7} m}=6.7 \cdot 10^{14} Hz$

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Which particle would have the slowest rate of deposition? A.round particle B.very large particle C.particle with sharp ends D.pa

lisabon 2012 [21]

The particle with sharp ends have the slowest rate of deposition

Answer: Option C

<u>Explanation:</u>

As per aerosol physics, deposition is a process where aerosol particles accumulate or settle on solid surfaces. Thereby, it reduces the concentration of particles in the air. Deposition velocity (rate of deposition) defines from F = vc, where v is deposition rate, F denotes flux density and c refers concentration.

Deposition velocity is slowest for particles of intermediate-sized particles because the frictional force offers resistance to the flow. Density is directly proportional to the deposition rate so clearly shows that high-density particles settle faster. Due to friction, round and large-sized particles deposit faster than oval/flattened sediments.

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3 years ago

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Which is the best procedure to make a permanent magnet?

notka56 [123]

placing a magnetically hard material in a strong magnetic field

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Which type of galaxy has the most active star formation?

UkoKoshka [18]

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Tectonic plates are large segments of the earth's crust that move slowly. Suppose one such plate has an average speed of 6.0 cm

faltersainse [42]

Answer:

1.35×10⁻⁷ m

37.278 mi/My

Explanation:

Speed of the tectonic plate= 6 cm/yr

Converting to seconds

$6=\frac{6}{365.25\times 24\times 60\times 60}$

So in one second it will move

$\frac{6}{365.25\times 24\times 60\times 60}$

In 71 seconds

$71\times \frac{6}{365.25\times 24\times 60\times 60}=1.35\times 10^{-5}\ cm$

The tectonic plate will move 1.35×10⁻⁵ cm or 1.35×10⁻⁷ m

Convert to mi/My

1 cm = 6.213×10⁻⁶ mi

1 M = 10⁶ years

$6\times 6.213\times 10^{-6}\times 10^6=37.278\ mi/My$

Speed of the tectonic plate is 37.278 mi/My

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3 years ago

Human reaction times are worsened by alcohol. How much further (in feet) would a drunk driver's car travel before he hits the br

sweet-ann [11.9K]

Answer:

A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes

Explanation:

Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:

2as = Vf² - Vi²

s = (Vf² - Vi²)/2a

where,

Vf = Final Velocity of Car = 0 mi/h

Vi = Initial Velocity of Car = 50 mi/h

a = deceleration of car

s = distance covered

Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.

So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:

s = vt

FOR SOBER DRIVER:

v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s

t = 0.33 s

s = s₁

Therefore,

s₁ = (73.33 ft/s)(0.33 s)

s₁ = 24.2 ft

FOR DRUNK DRIVER:

v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s

t = 1 s

s = s₂

Therefore,

s₂ = (73.33 ft/s)(1 s)

s₂ = 73.33 ft

Now, the distance traveled by drunk driver's car further than sober driver's car is given by:

ΔS = s₂ - s₁

ΔS = 73.33 ft - 24.2 ft

6 0

3 years ago