Answer: 4nmeter
Explanation: The two observer a and b will measure the same wavelength since the speed of the space craft is very small compared with the speed of light c. That is
V which is the speed of space craft 15000km/s = 15000000m/s
Comparing this with the speed of light c 3*EXP(8)m/s we have
15000000/300000000
= 0.05=0.1
Therefore the speed of the space craft V in terms of the speed of light c is 0.1c special relativity does not apply to object moving at such speed. So the wavelength would not be contracted it will remain same for both observers.
I think the key here is to be exquisitely careful at all times, and
any time we make any move, keep our units with it.
We're given two angular speeds, and we need to solve for a time.
Outer (slower) planet:
Angular speed = ω rad/sec
Time per unit angle = (1/ω) sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/ω sec/rad) · (2π rad) = 2π/ω seconds .
Inner (faster) planet:
Angular speed = 2ω rad/sec
Time per unit angle = (1/2ω) sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/2ω sec/rad) · (2π rad) = 2π/2ω sec = π/ω seconds.
So far so good. We have the outer planet taking 2π/ω seconds for one
complete revolution, and the inner planet doing it in only π/ω seconds ...
half the time for double the angular speed. Perfect !
At this point, I know what I'm thinking, but it's hard to explain.
I'm pretty sure that the planets are in line on the same side whenever the
total elapsed time is something like a common multiple of their periods.
What I mean is:
They're in line, SOMEwhere on the circles, when
(a fraction of one orbit) = (the same fraction of the other orbit)
AND
the total elapsed time is a common multiple of their periods.
Wait ! Ignore all of that. I'm doing a good job of confusing myself, and
probably you too. It may be simpler than that. (I hope so.) Throw away
those last few paragraphs.
The planets are in line again as soon as the faster one has 'lapped'
the slower one ... gone around one more time.
So, however many of the longer period have passed, ONE MORE
of the shorter period have passed. We're just looking for the Least
Common Multiple of the two periods.
K (2π/ω seconds) = (K+1) (π/ω seconds)
2Kπ/ω = Kπ/ω + π/ω
Subtract Kπ/ω : Kπ/ω = π/ω
Multiply by ω/π : K = 1
(Now I have a feeling that I have just finished re-inventing the wheel.)
And there we have it:
In the time it takes the slower planet to revolve once,
the faster planet revolves twice, and catches up with it.
It will be 2π/ω seconds before the planets line up again.
When they do, they are again in the same position as shown
in the drawing.
To describe it another way . . .
When Kanye has completed its first revolution ...
Bieber has made it halfway around.
Bieber is crawling the rest of the way to the starting point while ...
Kanye is doing another complete revolution.
Kanye laps Bieber just as they both reach the starting point ...
Bieber for the first time, Kanye for the second time.
You're welcome. The generous bounty of 5 points is very gracious,
and is appreciated. The warm cloudy water and green breadcrust
are also delicious.
<span>The oven will transform electrical energy into heat energy</span>
Answer:
Given:
Mass of the polar bear (m) = 6.8 kg
Speed of the polar bear (v) = 5.0 m/s
To Find:
Kinetic energy of the polar bear (KE)
Explanation:
Formula:
Substituting values of m & v in the equation:
Kinetic energy of the polar bear (KE) = 23002.1 J
Answer:
The workdone is 
Explanation:
From the question we are told that
The height of the cylinder is 
The face Area is 
The density of the cylinder is 
Where 
is the density of freshwater which has a constant value
Now
Let the final height of the device under the water be 
Let the initial volume underwater be 
Let the initial height under water be 
Let the final volume under water be 
According to the rule of floatation
The weight of the cylinder = Upward thrust
This is mathematically represented as
So 
=> 
Now the work done is mathematically represented as
![= \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.](https://tex.z-dn.net/?f=%3D%20%20%20%5Crho_w%20g%20A%20%5B%5Cfrac%7Bh%5E2%7D%7B2%7D%20%5D%20%5Cleft%20%7C%20h_f%7D%20%5Catop%20%7Bh%7D%7D%20%5Cright.)
![= \frac{g A \rho}{2} [h^2 - h_f^2]](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bg%20A%20%5Crho%7D%7B2%7D%20%20%5Bh%5E2%20-%20h_f%5E2%5D)
![= \frac{g A \rho}{2} (h^2) [1 - \frac{h_f^2}{h^2} ]](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bg%20A%20%5Crho%7D%7B2%7D%20%28h%5E2%29%20%20%5B1%20%20-%20%5Cfrac%7Bh_f%5E2%7D%7Bh%5E2%7D%20%5D)
Substituting values
