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3 years ago

How are wavelength and frequency of a wave related?

2 answers:

8 0

Wavelength is the length from one peak of a wave to the next, while frequency is the amount of waves present in a unit of time. The greater the wavelength the smaller the frequency. The smaller the wavelength, the bigger the frequency.

tester [92]3 years ago

6 0

The product of

(wavelength) x (frequency)

is always the same number. It's the speed of the wave.

So if either number changes, the other will always change by the
same factor in the opposite direction, in order to keep their product
constant.

You might be interested in

Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point,F⃗ 1F→1F_1_vec has a magnitude of 8.80 NN and is directed at an an

castortr0y [4]

Answer:

Fx = -9.15 N
Fy = 1.72 N
F∠γ ≈ 9.31∠-10.6°

Explanation:

You apparently want the sum of forces ...

F = 8.80∠-56° +7.00∠52.8°

Your angle reference is a bit unconventional, so we'll compute the components of the forces as ...

f∠α = (-f·cos(α), -f·sin(α))

This way, the 2nd quadrant angle that has a negative angle measure will have a positive y component.

= -8.80(cos(-56°), sin(-56°)) -7.00(cos(52.8°), sin(52.8°))

≈ (-4.92090, 7.29553) +(-4.23219, -5.57571)

≈ (-9.15309, 1.71982)

The resultant component forces are ...

Fx = -9.15 N
Fy = 1.72 N

Then the magnitude and direction of the resultant are

F∠γ = (√(9.15309² +1.71982²))∠arctan(-1.71982/9.15309)

F∠γ ≈ 9.31∠-10.6°

4 0

3 years ago

The moon Phobos orbits Mars

podryga [215]

Explanation:

For a circular orbit v= $\sqrt{\frac{G.m}{r} }$ with G = 6.6742 × $10^{-11}$

Given m = 6.42 x 10^23 kg and r=9.38 x 10^6 m

=> v = 2137.3 m/s

I hope this is the correct way to solve

3 0

2 years ago

1. Calculate the charge in a circuit which carries a current of 30 amperes for 5 seconds

sweet-ann [11.9K]

Answer:

1. 150C.

2. 50sec

3.1.5a

Explanation:

1. I = Q/T

Q= 30x5

=150c

2.applying the formulae, I = Q/T

T= Q/I

=500/10

=50sec.

3. using the formulae i=q/t

i= 120/80

=1.5a.

6 0

3 years ago

A felt-covered beanbag is fired into a empty wooden crate that sits on a concrete floor and is open on one side. The beanba

Phantasy [73]

Answer:

31.42383 m/s

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

$\mu$ = Coefficient of kinetic friction = 0.48

s = Displacement = 0.935 m

$m_1$ = Mass of bean bag = 0.354 kg

$m_2$ = Mass of empty crate = 3.77 kg

$v_1$ = Speed of the bean bag

$v_2$ = Speed of the crate

Acceleration

$a=-\frac{f}{m}\\\Rightarrow a=-\frac{\mu mg}{m}\\\Rightarrow a=-\mu g$

$a=--9.81\times 0.48=4.7088\ m/s^2$

From equation of motion

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.7088\times 0.935+0^2}\\\Rightarrow v=2.96739\ m/s$

In this system the momentum is conserved

$m_1v_1=(m_1+m_2)v_2\\\Rightarrow v_1=\frac{(m_1+m_2)v_2}{m_1}\\\Rightarrow u=\frac{(0.354+3.77)\times 2.69739}{0.354}\\\Rightarrow u=31.42383\ m/s$

The speed of the bean bag is 31.42383 m/s

8 0

3 years ago

In what order does light pass through structures of the eye? lens, cornea, retina cornea, pupil, lens pupil, iris, lens pupil, c

Mrrafil [7]

Answer: Light passes through the front of the eye (cornea) to the lens. The cornea and the lens help to focus the light rays onto the back of the eye (retina). The cells in the retina absorb and convert the light to electrochemical impulses which are transferred along the optic nerve and then to the brain.

Explanation:

When light rays reflect off an object and enter the eyes through the cornea (the transparent outer covering of the eye), you can then see that object. The cornea bends, or refracts, the rays that pass through the round hole of the pupil.

4 0

3 years ago