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3 years ago

Please help, I do not understand

Physics

1 answer:

Anettt [7]3 years ago

5 0

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.

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Answer:

The value of the average convection coefficient is 20 W/Km².

Explanation:

Given that,

For first object,

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$\dfrac{h_{1}L_{1}}{k_{1}}=\dfrac{h_{2}L_{2}}{k_{2}}$

Here, $k_{1}=k_{2}$

$h_{2}=h_{1}\times\dfrac{L_{1}}{L_{2}}$

$h_{2}=\dfrac{q}{T_{2}-T_{1}}\times\dfrac{L_{1}}{L_{2}}$

Put the value into the formula

$h_{2}=\dfrac{10000}{400-300}\times\dfrac{0.5}{2.5}$

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ikadub [295]

(a) $-39.4^{\circ}$

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$p_y =0$

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

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where

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$v_1= 4.60 m/s$ is the velocity of the cue ball after the collision

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$\phi_1=28.0^{\circ}$ is the angle of the cue ball with the x-axis

$\phi_2$ is the angle of the second ball

Solving for $\phi_2$ , we find the angle between the direction of motion of the second ball and the original direction of motion:

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To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

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The initial total momentum along the x-direction as

$p_x = m u$

where

m is the mass of the cue ball

$u$ is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

$mu = 6.69 m\\u = 6.69 m/s$

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3 years ago