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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That's the job of the tiny "hair cells", located in the <em>inner ear.</em>

If you're a sound wave, this is how you reach the hair cells:

-- go into the big funnel of skin on the outside of the head, that thing we call the "ear"

-- go about an inch or two, down through a skinny dark tunnel inside the skull

-- at the end of the tunnel, hit a dead end, made of a wall of thin skin like a drum, called the "ear drum"; sound waves hit the ear drum and make it vibrate

-- on the other side of the ear drum, inside, is the chamber called the "middle ear". In there are the three smallest bones in the body; the ear drum touches the first one and makes it vibrate; the first one touches the second one and makes it vibrate; the second one touches the third one and makes it vibrate; then the third one touches another dead end made of thin skin.

-- the region on the other side of this wall of thin skin is the "inner ear"; it's a long skinny chamber, called the "cochlea", wound up in a spiral and filled with liquid; the walls of the cochlea are lined with millions of tiny hairs, sticking out into the liquid; the vibrations make waves in the liquid, and the waves make the tiny hairs wave back and forth; each tiny hair is the end of a nerve that goes into the brain; when that hair wiggles, it sends a nerve "message" into the brain.

-- there are two complete copies of this whole structure ... one on each side of your head.

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Explanation:

An diagram of the seed's motion is attached to this solution.

This is very close to a projectile motion question. And the quantity to be calculated, how far along the grant a seed released would travel is called the Range.

And this would be obtained from the equations of motion,

First of, the height of the plant is related to some quantities of the motion with this relation.

H = u(y) t + 0.5g(t^2)

U(y) = initial vertical component of velocity = 0 m/s, H = height at which motion began, = 20cm = 0.2 m

That means t = √(2H/g)

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R = u(x) t + 0.5g(t^2) = u(x) t (the second part of the equation goes to zero as the vertical component of the acceleration of this motion is 0)

(substituting the t = √(2H/g) derived from above

R = u(x) √(2H/g)

Where u(x) = the initial horizontal component of the bomb's velocity = maximum initial speed, that is, 4.6 m/s, H = vertical height at which the seed was released = 20 cm = 0.2 m, g = acceleration due to gravity = 9.8 m/s2

R = 4.6 √(2×0.2/9.8) = 0.929 m = 0.93 m = 93 cm. Option B.

QED!