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3 years ago

What is the answer to life?

2 answers:

8 0

The answer to life is different for everyone. There are ways to help you find out your personal answer to life, but in the end it's only you who can answer that question. Maybe you find it in a religion, in your sweetheart and family, at work, in friends, in books, etc. For example me: The purpose to my life is to work hard now so I can provide a safe future for my future wife and children, raise them up as people of good, grow old with the love of my life, die, and wait for the Lord's Second Coming. But as I said the answer to your question depends on your goals for life.

taurus [48]3 years ago

6 0

The answer to life is how ever you make it...... you can do anything with life that you would like.................. But make your choices worth it......

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lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A

qaws [65]

$2.6×10^6\:\text{m}$

Explanation:

The acceleration due to gravity g is defined as

$g = G\dfrac{M}{R^2}$

and solving for R, we find that

$R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)$

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration $F_c$ experienced by the satellite is equal to the gravitational force $F_G$ or

$F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)$

The orbital velocity v is the velocity of the satellite around the planet defined as

$v = \dfrac{2\pi r}{T}$

where r is the radius of the satellite's orbit in meters and T is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

$\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}$

Solving for M, we get

$M = \dfrac{4\pi^2 r^3}{GT^2}$

Putting this expression back into Eqn(1), we get

$R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}$

$\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}$

$\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}$

$\:\:\:\:= 2.6×10^6\:\text{m}$

5 0

3 years ago

Find the separation of two points on the Moon’s surface that can just be resolved by the 200 in. (= 5.1 m) telescope at Mount Pa

rewona [7]

Answer:

The separation of the 2 points should be 50.0 meters.

Explanation:

According to Rayleigh's scattering criteria the angular separation between 2 points to be resolved equals

$\theta =1.22\cdot \frac{\lambda }{D}$

Applying the given values we get

$\theta =1.22\cdot \frac{550\times 10^{-9} }{5.1}=0.1316\times 10^{-6}rads$

thus the linear separation equals $L=\theta \times Distance$

Applying the given values we get

$L=0.1316\times 10^{-6} \times 3.8\times 10^{5}\times 10^{3}meters\\\\\therefore L=50.00metes$

6 0

3 years ago

Anyone please help thank you

Thepotemich [5.8K]

Answer:

A) earth

B) live

C) live

D) earth

Explanation:

hope i help

4 0

2 years ago

At a fixed point, P, the electric and magnetic field vectors in an electromagnetic wave oscillate at angular frequency w . At wh

inn [45]

Answer:

Poynting vector oscillate at a frequency of 2omega

Explanation:

This is because The poynting vector is proportional to the cross product of electric and magnetic field vectors. So Because both fields oscillate sinusoidally with frequency w, trigonometric identities show that their product is a sinusoidal function of frequency of 2w.

4 0

3 years ago

By how many decibels do you reduce the sound intensity level due to a source of sound if you triple your distance from it? assum

sweet-ann [11.9K]

New sound intensity level after the move is

      10 log (1/3²)

   = -20 log (3)

   = -20 (0.4771) =   9.54 dB LOWER than before the move.

4 0

3 years ago