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AlexFokin [52]
1 year ago
9

Given that the wavelengths of visible light range from 400 nm to 700 nm, what is the highest frequency of visible light? (ccc =

3.0 x 108 m/s)
Physics
1 answer:
enot [183]1 year ago
4 0

7.5 × 10¹⁴ Hz is the highest frequency of visible light when wavelengths of visible light range from 400 nm to 700 nm.

The distance a wave travels in one unit of time is known as the wave speed (v).Taking into account that the wave travels one wavelength in one interval,

v=λ/T

Given that T = 1/f, we can write the equation above as,

V = f λ

Given data:

Minimum wavelength of visible light = 400 nm = 4 × 10⁻⁷ m

Speed of light = 3 × 10⁸ m/s

Frequency = c/λ = 3 × 10⁸ / 4 × 10⁻⁷

= 7.5 × 10¹⁴ Hz

To learn more about Frequency: brainly.com/question/16200748

#SPJ4

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A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1 2 and t = 1 s? Use G
Romashka-Z-Leto [24]

There is one mistake in the question.The Correct question is here

A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1/2 and t = 1 s? Use Galileo's formula v(t) = −9.8t m/s.

Answer:

y(1s) - y(1/2s) =  - 3.675 m  

The cat falls 3.675 m between time 1/2 s and 1 s.

Explanation:

Given data

time=1/2 sec to 1 sec

v(t)=-9.8t m/s

To find

Distance

Solution

As the acceleration as first derivative of velocity with respect to time  

So

acceleration(-g)=  dv/dt

Solve it

dv  =  a dt

dv =  -g dt

v - v₀  =  -gt

v=  dy/dt

dy  =  v dt

dy =  ( v₀ - gt ) dt

y(1s) - y(1/2s)  =  ( v₀ ) ( 1 - 1/2 ) - ( g/2 )[ ( t1)² -( t1/2s )² ]

y(1s) - y(1/2s)  = ( - 9.8/2 ) [ ( 1 )² - ( 1/2 )² ]

y1s - y1/2s  = ( - 4.9 m/s² ) ( 3/4 s² )

y(1s) - y(1/2s) =  - 3.675 m  

The cat falls 3.675 m between time 1/2 s and 1 s.

6 0
3 years ago
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Answer:

I would believe that it would be the last option

Explanation:

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4 0
2 years ago
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A = 94.22 Newtons

b = 58.16 kg

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3 years ago
A 20kg object acceleration by a force of 200N with coefficient of kineticfriction is 0.4 what is acceleration of the object?​
Schach [20]

Answer:

<u>Given</u><em> </em><em>-</em><em> </em><u>M</u><u> </u><u>=</u><u> </u>20 kg

k = 0.4

F = 200 N

<u>To </u><u>find </u><u>-</u><u> </u> acceleration

<u>Solution </u><u>-</u><u> </u>

F= kMA

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2 years ago
A team exerts a force of 10 N towards the south in a tug of war. The other, opposite team, exerts a force of 17 N towards the no
ikadub [295]

Answer:

Option C

Explanation:

According to the question:

Force exerted by the team towards south, F = 10 N

Force exerted by the opposite team towards North, F' = 17 N

Net Force, \vec{F_{net}} = \vec{F'} - \vec{F}

\vec{F_{net}} = \vec{F'} - \vec{F} = 17 - 10 = 7 N

Thus the force will be along the direction of force whose magnitude is higher

Therefore,

\vec{F_{net}} = 7 N towards North

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3 years ago
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