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2 years ago

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Kepler's laws, satellites motion and weightlessness

1 answer:

Anon25 [30]2 years ago

3 0

Answer:

First Kepler law states that <em><u>Each</u></em><em><u> </u></em><em><u>planet</u></em><em><u> </u></em><em><u>describes</u></em><em><u> </u></em><em><u>an</u></em><em><u> </u></em><em><u>ellipsoidal</u></em><em><u> </u></em><em><u>motion</u></em><em><u> </u></em><em><u>about</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>sun</u></em><em><u> </u></em><em><u>as</u></em><em><u> </u></em><em><u>its</u></em><em><u> </u></em><em><u>single</u></em><em><u> </u></em><em><u>focus</u></em><em><u>.</u></em>

Second Kepler law states that <em><u>A</u></em><em><u>n</u></em><em><u> </u></em><em><u>i</u></em><em><u>m</u></em><em><u>a</u></em><em><u>g</u></em><em><u>i</u></em><em><u>n</u></em><em><u>a</u></em><em><u>r</u></em><em><u>y</u></em><em><u> </u></em><em><u>l</u></em><em><u>i</u></em><em><u>n</u></em><em><u>e</u></em><em><u> </u></em><em><u>j</u></em><em><u>o</u></em><em><u>i</u></em><em><u>n</u></em><em><u>i</u></em><em><u>n</u></em><em><u>g</u></em><em><u> </u></em><em><u>a</u></em><em><u> </u></em><em><u>planet</u></em><em><u> </u></em><em><u>t</u></em><em><u>o</u></em><em><u> </u></em><em><u>t</u></em><em><u>h</u></em><em><u>e</u></em><em><u> </u></em><em><u>Sun</u></em><em><u> </u></em><em><u>sweeps</u></em><em><u> </u></em><em><u>out</u></em><em><u> </u></em><em><u>equal</u></em><em><u> </u></em><em><u>areas</u></em><em><u> </u></em><em><u>in</u></em><em><u> </u></em><em><u>equal</u></em><em><u> </u></em><em><u>time</u></em><em><u> </u></em><em><u>intervals</u></em><em><u>.</u></em>

Third Kepler law states that <em><u>The</u></em><em><u> </u></em><em><u>squares</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>period</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>revolution</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>planet</u></em><em><u> </u></em><em><u>around</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>sun</u></em><em><u> </u></em><em><u>are</u></em><em><u> </u></em><em><u>proportional</u></em><em><u> </u></em><em><u>to</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>cubes</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>mean</u></em><em><u> </u></em><em><u>distance</u></em><em><u> </u></em><em><u>between</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>planet</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>sun</u></em><em><u>.</u></em>

Weightlessness is the condition where the body has zero gravity ( its acceleration is equal to the acceleration due to gravity )

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MatroZZZ [7]

Answer:

The work done by the frictional force is 600J.

Explanation:

The work $W$ done by the frictional force is

$W= Fd$ .

Now, $F = 60N$ and $d =10m$ ; therefore,

$W= (60N)(10m)$

$\boxed{W = 600J.}$

Hence, the work done by friction is 660J.

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3 years ago

What are the three longest wavelengths for standing waves on a 264- cm -long string that is fixed at both ends

miv72 [106K]

The three longest wavelengths for the standing waves on a 264-cm long string that is fixed at both ends are:

5.2 meters.
2.6 meters.
1.7meters.

Given data:

Length of the fixed string = 264cms = 2.64 meters

The wavelength for standing waves is given by:

λ = 2L/n

where,

λ is the wavelength
L is the length of the string

For n = 1,

λ = 2×2.6/1

= 5.2 meters

For n = 2,

λ = 2×2.6/2

= 2.6 meters

For n = 3,

λ = 2×2.6/3

= 1.7 meters

To learn more about standing waves: brainly.com/question/14151246

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2 years ago

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3 years ago

Red laser light from a He-Ne laser (λ = 632.8 nm) creates a second-order fringe at 53.2° after passing through the grating. What

Svetlanka [38]

Explanation:

It is given that,

Wavelength of red laser light, $\lambda=632.8\ nm=632.8\times 10^{-9}\ m$

The second order fringe is formed at an angle of, $\theta=53.2^{\circ}$

For diffraction grating,

$d\ sin\theta=n\lambda$

$d=\dfrac{n\lambda}{sin\theta}$ , n = 2

$d=\dfrac{2\times 632.8\times 10^{-9}}{sin(53.2)}$

$d=1.58\times 10^{-6}\ m$

The wavelength λ of light that creates a first-order fringe at 22 is given by :

$\lambda=d\ sin\theta$

$\lambda=1.58\times 10^{-6}\ sin(22)$

$\lambda=5.91\times 10^{-7}\ m$

$\lambda=591\ nm$

Hence, this is the required solution.

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3 years ago

I started to solve this problem but I’m not sure on what to do next.

snow_tiger [21]

ANSWER:

3408.81 kg

STEP-BY-STEP EXPLANATION:

Given:

v = 111 m/s

Ek = 21000000 J

We have that the formula for kinetic energy is as follows:

$E_k=\frac{1}{2}\cdot m\cdot v^2$

We substitute the values given in the exercise and solve for m (mass)

$\begin{gathered} 21000000=\frac{1}{2}\cdot m\cdot111^2 \\ m=\frac{21000000\cdot2}{111^2} \\ m=3408.81\text{ kg} \end{gathered}$

The mass of the helicopter is 3408.81 kilograms.

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1 year ago