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3 years ago

HELP PLS

1 answer:

5 0

There are a variety of waves from light waves to mechanical waves. Waves can exhibit different effects like the Doppler Effect.

All light waves behave in a similar manner. They either get transmitted, reflected, absorbed, refracted, polarized, diffracted, or scattered based off of the composition of the object and the wavelength of the light.

According to Wikipedia, “One important property of mechanical waves is that their amplitudes are measured in an unusual way, displacement divided by (reduced) wavelength. When this gets comparable to unity, significant nonlinear effects such as harmonic generation may occur, and, if large enough, may result in chaotic effects.” Mechanical waves are chaotic and its “amplitudes” are measured unusually.

Diffraction is when light bends around objects and spread after passing out through small openings. “Diffraction occurs with all waves, including sound waves, water waves, and electromagnetic waves such as light that the eye can see.”-Wikipedia. Here is the formula to Diffraction: d sin θ = nλ

Doppler effect can occur for any type of wave like sound or water waves. An example of this is when we hear a police car with its sirens on, coming towards us. The closer you are to the police car, the higher the wavelength, but the farther away you are, the lower the wavelength.

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Un the way to the moon, the Apollo astro-

kherson [118]

Answer:

Distance = 345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

Second part

The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

$a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2} } \\a=3.33*10^{19} [m/s^2]$

6 0

3 years ago

Explain how does a convex lens produce real image from virtual object

Sphinxa [80]

Answer:

Real images are formed where rays of light actually converge, whereas virtual images occur with they are perceived to converge. Real images can be produced by passing light through converging lenses or with a concave mirror of some sort.

Explanation:

hope that helped a little :)

6 0

3 years ago

Read 2 more answers

Which element(s) is/are not balanced in this equation?

Shkiper50 [21]

All elements are balanced. There are 1 Mg, 1 O, 2 Li's and 2 Cl's.

3 0

3 years ago

How we can best share the observation of the light through one hole the paper

zhenek [66]