Answer:
R (120) = 940Ω
Explanation:
The variation in resistance with temperature is linear in metals
ΔR (T) = R₀ α ΔT
where α is the coefficient of variation of resistance with temperature, in this case α = -0,0005 / ºC
let's calculate
ΔR = 1000 (-0,0005) (120-0)
ΔR = -60
Ω
ΔR = R (120) + R (0) = -60
R (120) = -60 + R (0)
R (120) = -60 + 1000
R (120) = 940Ω
Answer:
<em>Velocity is the rate at which the position changes</em>
<em>Velocity is the rate at which the position changesWhy do we need</em>
<em>Velocity is the rate at which the position changesWhy do we needVectors make it convenient to handle quantities going in different directions</em><em>.</em><em>.</em><em> </em>
Explanation:
Thank you!
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.
Answer:
(a) <u>11.3 L</u>
(b) <u>10 M</u>
Explanation:
The mass-luminosity relationship states that:
Luminosity ∝ Mass^3.5
Luminosity = (Constant)(Mass)^3.5
So, in order to find the values of luminosity or mass of different stars, we take the luminosity or mass of sun as reference. Therefore, write the equation for a star and Sun, and divide them to get:
Luminosity of a star/L = (Mass of Star/M)^3.5 ______ eqn(1)
where,
L = Luminosity of Sun
M = mass of Sun
(a)
It is given that:
Mass of Star = 2M
Therefore, eqn (1) implies that:
Luminosity of star/L = (2M/M)^3.5
Luminosity of Star = (2)^3.5 L
<u>Luminosity of Star = 11.3 L</u>
(b)
It is given that:
Luminosity of star = 3160 L
Therefore, eqn (1) implies that:
3160L/L = (Mass of Star/M)^3.5
taking ln on both sides:
ln (3160) = 3.5 ln(Mass of Star/M)
8.0583/3.5 = ln(Mass of Star/M)
Mass of Star/M = e^2.302
<u>Mass of Star = 10 M</u>
Answer:
mass × gravity ×height. so....that.. Ep = mgh. Ep= 35×9.8×5. Ep = 1715.
Explanation: