Why do not we observe space quantization for spinning top?

Suppose a small planet is discovered that is 16 times as far from the Sun as the Earth's distance is from the Sun. Use Kepler's

mamaluj [8]

Answer:

23376 days

Explanation:

The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

$T^2\alpha R^3\\T^2=kR^3.......................(1)$

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

$\frac{T^2}{R^3}=k.......................(2)$

Let the orbital period of the earth be $T_e$ and its mean distance of from the sun be $R_e$ .

Also let the orbital period of the planet be $T_p$ and its mean distance from the sun be $R_p$ .

Equation (2) therefore implies the following;

$\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)$

We make the period of the planet $T_p$ the subject of formula as follows;

$T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)$

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

$R_p=16R_e...............(5)$

Substituting equation (5) into (4), we obtain the following;

$T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}$

$R_e^3$ cancels out and we are left with the following;

$T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)$

Recall that the orbital period of the earth is about 365.25 days, hence;

$T_p=64*365.25\\T_p=23376days$

4 0

2 years ago

The nonreflective coating on a camera lens with an index of refraction of 1.29 is designed to minimize the reflection of 634-nm

eimsori [14]

Answer:

minimum thickness of the coating = 122.868 nm

Explanation:

Given data

lens index of refraction = 1.29

wavelength = 634 nm

glass index of refraction = 1.53

to find out

minimum thickness of the coating

solution

we have given non reflective coating

so

we know that minimum thickness of the coating formula

minimum thickness of the coating = Wavelength / 4n

here n is coating index of refraction

so put here both value to get thickness

minimum thickness of the coating = Wavelength / 4n

minimum thickness of the coating = 634 / 4 ( 1.29 )

so minimum thickness of the coating = 122.868 nm

5 0

2 years ago

If I bought 59 kids off the black market in a secret door in the stalls and I sell 21 kids then ask for 81 more then 12 off the

larisa [96]

You will be left with 106 kids

<h3>Meaning of word problem</h3>

A word problem can be defined as a mathematical problem that is written in word or written in a sentence format.

In a word problem, the student is expected to decode the sentence into a mathematical expression before solving

In conclusion, You will be left with 106 kids

Learn more about word problems: brainly.com/question/13818690

#SPJ1

5 0

1 year ago

A ball is thrown straight up from ground level. It passes a 2-m-high window. The bottom of the window is 7.5 m off the ground. T

slamgirl [31]

Answer:

$u=14.48m/s$

Explanation:

From the question we are told that:

Height of window $h=2m$

Height of window off the ground $h_g=7.5m$

Time to fall and drop $t=1.3s$

Generally the Newton's equation motion is mathematically given by

$s=ut+\frac{1}{2}at^2$

Where

$h=ut+\frac{1}{2}at^2$

$2=u1.3-\frac{1}{2}*9.8*1.3^2$

$2=u1.3-8.281$

$u=7.91m/s^2$

Generally the Newton's equation motion is mathematically given by

$2as=v^2-u^2$

Where

$-2gh_g=v^2-u^2$

$-2*9.8*7.5=(7.91)^2-u^2$

$-147=62.5681-u^2$

$u=\sqrt{209.5681}$

$u=14.48m/s$

Therefore the ball’s initial speed

$u=14.48m/s$

8 0

2 years ago

Which is a chemical change?

statuscvo [17]

Answer:

A. Splitting water into hydrogen and oxygen.

Explanation:

A chemical change produces a new substance. ... A physical change does not produce a new substance. Changes in state or phase (melting, freezing, vaporization, condensation, sublimation) are physical changes.

8 0

2 years ago

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