What is the process called when the moon begins to fade from a full moon to the new moon?

Visible light waves do not diffract as well as radio waves because

Anvisha [2.4K]

they have more energy than radio waves.

&

because the wavelength of the light waves are too small

3 0

3 years ago

If you shine a light of frequency 375hz on a double slit setup , and you measure the slit separation to be 950 nm and the screen

amm1812

Answer:

y = 33.93 10⁵ m

Explanation:

This is an interference exercise, for the contributory interference is described by the expression

d sin θ = m λ

let's use trigonometry for the angle

tan θ = y / L

how the angles are small

tan θ = sin θ / cos tea = sin θ

we substitute

sin θ = y / L

d y / L = m λ

y = m λ L / d

the light fulfills the relation of the waves

c = λ f

λ = c / f

λ = 3 10⁸ /375

λ = 8 10⁵ m

first order m = 1

let's calculate

y = 1 8 10⁵ 4030 10-9 / 950 10-9

y = 33.93 10⁵ m

6 0

3 years ago

Determine the slope of end a of the cantilevered beam. E = 200 gpa and i = 65. 0(106) mm4

DENIUS [597]

For E = 200 gpa and i = 65. 0(106) mm4, the slope of end a of the cantilevered beam is mathematically given as

A=0.0048rads

<h3>What is the slope of end a of the cantilevered beam?</h3>

Generally, the equation for the is mathematically given as

$A=\frac{PL^2}{2EI}+\frac{ML}{EI}$

Therefore

A=\frac{10+10^2+3^2}{2*240*10^9*65*10^6}+\frac{10+10^3*3}{240*10^9*65*10^{-6}}

A=0.00288+0.00192=0.0048rads

A=0.0048rads

In conclusion, the slope is

A=0.0048rads

Read more about Graph

brainly.com/question/14375099

5 0

2 years ago

What is serve in volleyball?

madreJ [45]

Underhand serve is a type of volleyball serve in which a player holds the ball with one hand and swings the other hand in an arc motion, striking under the ball with a fist to put it into play. An underhand serve is the most common serve for beginners.

3 0

3 years ago

Read 2 more answers

A cat dozes on a stationary merry-go-round, at a radius of 5.5 m from the center of the ride. Then the operator turns on the rid

bixtya [17]

Answer:

$u_{s}=0.56$

Explanation:

For the cat to stay in place on the merry go round without sliding the magnitude of maximum static friction must be equal to magnitude of centripetal force

$F_{s.max}=\frac{mv^2}{r} \\$

Where the r is the radius of merry-go-round and v is the tangential speed

but

$F_{s.max}=u_{s}F_{N}=u_{s}mg$

So we have

$u_{s}mg=\frac{mv^2}{r}\\ u_{s}=\frac{v^2}{gr}\\ Where\\v=\frac{2\pi R}{T} \\So\\u_{s}=\frac{(\frac{2\pi R}{T} )^2}{gr} \\u_{s}=\frac{4\pi^2 r}{gT^2}$

Substitute the given values

So

$u_{s}=\frac{4\pi^2 5.5m}{(9.8m/s^2)(6.3s)^2} \\u_{s}=0.56$

6 0

3 years ago